ornia 
tal 

y 


A  TEXTBOOK 


ON 


STEAM  ENGINEERING 


INTERNATIONAL  CORRESPONDENCE  SCHOOLS 
SCRANTON,  PA. 


ARITHMETIC 

MENSURATION  AND  USE  OF  LETTERS  IN  FORMULAS 
PRINCIPLES  OF  MECHANICS 

MACHINE  ELEMENTS 

MECHANICS  OF  FLUIDS 

STRENGTH  OF  MATERIALS 


1048! 


SCRANTON 

INTERNATIONAL  TEXTBOOK  COMPANY 

A-2 


Copyright,  1897,  by  THE  COLLIERY  ENGINEER  COMPANY. 
Copyright,  1902,  by  INTERNATIONAL  TEXTBOOK  COMPANY. 


Entered  at  Stationers'  Hall,  London. 


Arithmetic :  Copyright,  1893,  1894,  1895, 1896, 1897,  1898, 1899,  by  THE  COLLIERY  EN- 
GINEER COMPANY.  Copyright,  1901,  by  INTERNATIONAL  TEXTBOOK  COMPANY. 

Mensuration  and  Use  of  Letters  in  Formulas :  Copyright,  1894,  1895,  1897, 1898,  by 
THE  COLLIERY  ENGINEER  COMPANY. 

Principles  of  Mechanics:  Copyright,  1901,  by  INTERNATIONAL  TEXTBOOK  COM- 
PANY. Entered  at  Stationers'  Hall,  London. 

Machine  Elements:  Copyright,  1901,  by  INTERNATIONAL  TEXTBOOK  COMPANY. 
Entered  at  Stationers'  Hall,  London. 

Mechanics  of  Fluids  :  Copyright,  1901,  by  INTERNATIONAL  TEXTBOOK  COMPANY. 
Entered  at  Stationers'  Hall.  London. 

Strength  of  Materials  :  Copyright,  1901.  by  INTERNATIONAL  TEXTBOOK  COMPANY. 
Entered  at  Stationers'  Hall,  London. 

Arithmetic,  Key  :  Copyright,  1893,  1894.  1896,  1897,  1898,  by  THE  COLLIERY  ENGI- 
NEER COMPANY. 

Mensuration  and  Use  of  Letters  in  Formulas,  Key  :  Copyright,  1894,  1895,  by  THE 
COLLIERY  ENGINEER  COMPANY. 


All  rights  reserved. 


Cl 

BURR   PRINTING  HOUSE, 

FRANKFORT  AND   JACOB  STREETS, 

NEW  YORK. 


715 


PREFACE. 


In  preparing  this  set  of  textbooks,  which  replaces  those 
formerly  sent  to  students  of  our  Stationary  Engineering 
Course  under  title  of  "The  Elements  of  Steam  Engineer- 
ing," we  have  aimed  to  produce  a  practical  treatise  on  the 
care,  operation,  and  management  of  all  kinds  of  steam 
plants,  with  the  result  that  these  volumes  form  the  most 
thorough,  practical,  and  comprehensive  treatise  on  the  sub- 
ject that  has  yet  been  published.  While  these  textbooks 
contain  much  that  is  of  direct  value  to  the  designer,  they 
have  been  prepared  more  particularly  to  meet  the  wants  of 
those  engaged  in  actual  work  in  steam  plants  and  of  those 
desiring  to  prepare  themselves  to  pass  examinations  for 
license. 

The  subject  of  steam  boilers  and  the  various  appliances 
found  in  boiler  rooms  has  been  very  thoroughly  covered. 
Enough  of  the  theory  has  been  given  to  enable  the  fireman 
or  engineer  to  surmount  the  usual  difficulties  encountered 
and  to  select  boilers  and  appliances  best  suited  to  particular 
conditions;  the  best  modern  practice  of  installing,  managing, 
and  testing  boiler  plants  is  also  given. 

Steam  engines  and  their  appliances,  and  condensers,  are 
treated  in  the  same  thorough  manner  as  steam  boilers.  The 
general  principles  underlying  the  operation  of  all  makes  of 
steam  engines,  engine  appliances,  and  condensers  are  so 
treated  that  the  engineer  will  be  enabled  not  only  to  satis- 
factorily and  economically  run  a  particular  engine,  but  will 
also  be  able  to  rapidly  trace  out  the  peculiarities  of  other 
iii 


iv  PREFACE. 

engines,  make  adjustments  of  all  kinds  of  governors  and 
valve  gears,  make  engine  tests,  take  and  read  indicator  dia- 
grams, and  select  engines  and  appliances  to  suit  existing 
conditions. 

The  sections  relating  to  pumps  are  especially  valuable  to 
pump  runners  and  engineers  in  charge  of  water-works 
plants.  Our  thanks  are  due  to  the  different  pump  manu- 
facturers, notably  to  Henry  R.  Worthington,  New  York, 
for  information  furnished  and  courtesies  extended. 

On  account  of  the  rapid  and  growing  increase  in  the  use 
of  elevators  in  large  buildings,  we  have  devoted  a  large  part 
of  one  volume  to  this  subject.  The  majority  of  elevators 
in  use  are  looked  after  by  the  engineers  in  charge  of  the 
buildings  in  which  they  are  located,  and  a  knowledge  of 
their  construction,  care,  and  management  is  an  invaluable 
aid  to  their  successful  operation.  The  present  is  the  first 
attempt  to  treat  all  existing  types  of  elevators,  and  we 
record  with  pleasure  that  in  preparing  this  portion  of  the 
Course  we  have  had  the  hearty  cooperation  of  the  following 
firms:  Otis  Elevator  Company  of  New  York;  Otis  Elevator 
Company  of  Chicago  (formerly  Crane  Elevator  Company) ; 
Morse,  Williams  &  Company,  Philadelphia  ;  A.  B.  See 
Manufacturing  Company,  Brooklyn;  The  Elektron  Manu- 
facturing Company,  Springfield,  Mass. ;  Burdett-Rowntree 
Manufacturing  Company,  Chicago;  Automatic  Switch  Com- 
pany, Baltimore,  Md. ;  The  Winslow  Brothers  Company, 
Chicago;  The  Mason  Regulator  Company,  Boston,  Mass.; 
The  Cutler-Hammer  Manufacturing  Company,  Milwaukee, 
Wis. ;  Sprague  Elevator  Company,  New  York;  The  Plunger 
Elevator  Company,  Worcester,  Mass. ;  Elevator  Supply  and 
Repair  Company,  New  York  and  Chicago.  Special  thanks 
are  due  to  The  Electrical  World  and  Engineer  for  infor- 
mation gleaned  from  its  files,  and  to  Mr.  R.  C.  Smith  and 
Mr.  E.  W.  Marshall  of  the  Otis  Elevator  Company  for 
assistance  rendered  in  the  preparation  of  this  treatise.  By 
courtesy  of  the  Otis  Elevator  Company  we  have  incorpo- 
rated a  few  valuable  articles  on  elevator  management  appear- 
ing in  their  pamphlet  "  Information  for  Engineers." 


PREFACE.  v 

The  paper  on  steam  heating  will  be  valuable  to  those 
engineers  who  look  after  the  steam-heating  systems  in  large 
buildings  and  also  to  those  who  desire  to  use  an  existing 
boiler  plant  to  heat  buildings  with  steam. 

On  account  of  the  inconvenience  resulting  from  the  large 
size  of  the  former  drawing  volume,  it  has  been  especially 
reset  for  this  edition  and  the  pages  made  uniform  with  the 
other  volumes.  On  account  of  its  thinness,  it  has  been 
combined  with  one  of  the  other  volumes,  thus  making  five 
volumes  in  the  set  instead  of  six,  as  formerly.  The  paper 
on  Mechanical  Drawing  has  been  thoroughly  revised  and 
much  new  matter  added. 

Among  those  who  assisted  in  the  preparation  of  these 
volumes,  in  addition  to  members  of  our  own  staff, 
were  Charles  J.  Mason,  Chief  Engineer  in  charge  of  the 
State  Street  power  station  of  the  Brooklyn  Rapid  Transit 
Company,  Brooklyn,  N.  Y. ,  and  L.  Hollingsworth,  Jr., 
formerly  Chief  Draftsman  of  the  Dickson  Manufacturing 
Company,  Scranton,  Pa.,  and  now  Chief  Draftsman  of  the 
Pennsylvania  Iron  Works  Company,  Philadelphia,  Pa.  The 
whole  Course  was  under  the  editorial  supervision  of  John 
A.  Grening. 

The  method  of  numbering  the  pages,  cuts,  articles,  etc., 
is  such  that  each  paper  and  part  is  complete  in  itself;  hence, 
in  order  to  make  the  indexes  intelligible  it  was  necessary  to 
give  each  paper  and  part  a  number.  This  number  is  placed 
at  the  top  of  each  page,  on  the  headline,  opposite  the  page 
number;  to  distinguish  it  from  the  page  number  it  is  pre- 
ceded by  the  printer's  section  mark  (§).  Consequently,  a 
reference  such  as  §  7,  page  28,  is  readily  found  as  follows: 
Look  along  the  inside  edges  of  the  headlines  until  §  7  is 
found,  and  then  through  §  7  until  page  28  is  found. 

The  Examination  Questions  and  their  answers  are 
grouped  together  at  the  ends  of  the  volumes  containing  the 
papers  to  which  they  refer,  and  are  given  the  same  section 
numbers. 

INTERNATIONAL  CORRESPONDENCE  SCHOOLS. 


CONTENTS. 


ARITHMETIC.                                                           S> 
Definitions 

action. 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
1 
2 
2 
2 
2 
2 
2 
2 
2 
2 

Page. 
I 

I 
4 
9 
11 
16 
20 
23 
25 
28 
31 
33 
37 
39 
41 
42 
44 
52 
1 
4 
9 
10 
13 
16 
18 
20 
21 

Notation  and  Numeration 

Addition    

Subtraction 

Multiplication     
Division      ...                

Cancelation     
Fractions 

Reduction  of  Fractions     
Addition  of  Fractions  
Multiplication  of  Fractions   
Division  of  Fractions   
Decimals 

Addition  of  Decimals 

Subtraction  of  Decimals        .           ... 

Multiplication  of  Decimals    
Division  of  Decimals                               . 

Signs  of  Aggregation  

Percentage      

Calculations  Involving  Percentage      .     . 
Denominate  Numbers 

Measures    

Reduction  of  Denominate  Numbers    . 
Addition  of  Denominate  Numbers 
Subtraction  of  Denominate  Numbers 
Multiplication  of  Denominate  Numbers 
Division  of  Denominate  Numbers  XV.     . 
vii 

viii  CONTENTS. 

ARITHMETIC — Continued.                                    Section.  Page. 

Involution 2  25 

Evolution 2  27 

Square  Root .     .     .  2  28 

Cube  Root 2  35 

Ratio 2  40 

Proportion 2  43 

Unit  Method 2  48 

MENSURATION  AND  USE  OF  LETTERS  IN  FORMULAS. 

Formulas 3  1 

Mensuration  ..." 3  13 

Quadrilaterals 3  17 

The  Triangle 3  20 

Polygons 3  24 

The  Prism  and  Cylinder 3  32 

The  Pyramid  and  Cone 3  36 

The  Frustum  of  a  Pyramid  or  Cone  .     .  3  37 

The  Sphere  and  Cylindrical  Ring  ...  3  39 

PRINCIPLES  OF  MECHANICS. 

Matter  and  Its  Properties 4  1 

Motion  and  Velocity 4  5 

Force 4  10 

Gravitation  and  Weight   ......  4  15 

Work,  Power,  and  Energy 4  20 

Composition  and  Resolution  of  Forces    .  4  26 

Friction      ...........  4  33 

Center  of  Gravity    .     .    • 4  39 

Centrifugal  Force    ........  4  43 

Equilibrium 4  45 

MACHINE  ELEMENTS. 

The  Lever,  Wheel,  and  Axle     ....  5  1 

Pulleys •  .     .     .  5  7 

Belts      . 5  12 

Wheel  Work 5  25 

Gear  Calculations    .  5  34 


CONTENTS. 


x 


MACHINE  ELEMENTS  —  Continued.                       Section.  Page. 

Fixed  and  Movable  Pulleys  .....       5  42 

The  Inclined  Plane       .......        5  47 

The  Screw      ..........       5  53 

Velocity  Ratio  and  Efficiency    ....       5  59 

MECHANICS  OF  FLUIDS. 

Hydrostatics  ........  ,  .     .  6  1 

Specific  Gravity       ........  6  15 

Buoyant  Effects  of  Water     .....  6  19 

Hydrokinetics     .........  6  22 

Pneumatics    ..........  6  27 

Pneumatic  Machines    .......  6  40 

Pumps  ............  6  47 

STRENGTH  OF  MATERIALS. 

General  Principles  ........        7  1 

Tensile  Strength  of  Materials   ....       7  2 

Crushing  Strength  of  Materials     ...        7  15 

Transverse  Strength  of  Materials  7  23 

Shearing,  or  Cutting,  Strength  of  Mate- 

rials ............       7  29 

Torsion      ...........       7  32 

EXAMINATION  QUESTIONS.  Section. 

Arithmetic,  Part  1  .......    ,.  1 

Arithmetic,  Part  2  ........  1 

Arithmetic,  Part  3  ........  1 

Arithmetic,  Part  4  .     .......  2 

Arithmetic,  Part  5  ........  2 

Mensuration  and  Use  of  Letters  in  For- 

mulas     ...........  3 

Principles  of  Mechanics    ......  4 

Machine  Elements  ........  5 

Mechanics  of  Fluids     .  .....  6 

Strength  of  Materials  .......  7 

ANSWERS  TO  EXAMINATION  QUESTIONS.  Section. 

Arithmetic,  Part  1  ........  1 

Arithmetic,  Part  2  .  1 


x  CONTENTS. 

ANSWERS  TO  EXAMINATION  QUESTIONS — Con- 
tinued. Section. 

Arithmetic,  Part  3 1 

Arithmetic,  Part  4 .    '  2 

Arithmetic,  Part  5  .     .    ".'•..    .     *     .     .  2 
Mensuration  and  Use  of  Letters  in  For- 
mulas     .     .     .     .     .     .     .     v     ...  3 

Principles  of  Mechanics 4 

Machine  Elements  ........  5 

Mechanics  of  Fluids »  G 

Strength  of  Materials  .  7 


ARITHMETIC. 

(PART  1.) 


DEFINITIONS. 

1.  Arithmetic  is  the  art  of  reckoning,  or  the  study  of 
numbers. 

2.  A  unit  is  one,  or  a  single  thing,  as  one,  one  bolt,  one 
•pulley,  one  dozen. 

3.  A  number  is  a  unit,  or  a  collection  of  units,  as  one^ 
three  engines,  five  boilers. 

4.  The  unit  of  a  number  is  one  of  the  collection  of  units 
which  constitutes  the  number.     Thus,  the  unit  of  twelve  is 
one,  of  twenty  dollars  is  one  dollar,  of  one  hundred  bolts  is 
one  bolt. 

5.  A   concrete   number  is  a  number  applied  to  some 
particular  kind  of  object  or  quantity,  as  three  grate  bars, 
five  dollars,  ten.  pounds. 

6.  An  abstract  number  is  a  number  that  is  not  applied 
to  any  object  or  quantity,  as  three,  five,  ten. 

7.  Iiike  numbers  are  numbers  which  express  units  of  the 
same  kind,  as  six  days  and  ten  days,  two  feet  and  five  feet. 

8.  Unlike  numbers  are  numbers  which  express  units  of 
different  kinds,  as  ten  months  and  eight  miles,  seven  wrenches 
and  five  bolts. 

NOTATION  AND  NUMERATION. 

9.  Numbers  are  expressed  in  three  ways:     (1)  by  words; 
(2)  by  figures;  (3)  by  letters. 

10.  Notation  is  the  art  of  expressing  numbers  by  figures 
or  letters. 

11.  Numeration  is  the  art  of  reading  the  numbers  which 
have  been  expressed  by  figures  or  letters. 

§1 

For  notice  of  copyright,  see  page  immediately  following  the  title  page. 


2  ARITHMETIC.  §  1 

12.  The  Arabic  notation  is  the  method  of  expressing 
numbers  by  figures.     This  method   employs   ten   different 
figures  to  represent  numbers,  viz. : 

Figures       0123456789 
naught,  one    two  three  four  five     six  seven  eight  nine 

Names  cipher, 
or  zero. 

The  first  character  (0)  is  called  naught,  cipher,  or  zero, 

and  when  standing  alone  has  no  value. 

The  other  nine  figures  are  called  digits,  and  each  has  a 
value  of  its  own. 

Any  whole  number  is  called  an  Integer. 

13.  As   there  are  only  ten  figures   used  in    expressing 
numbers,  each  figure  must  have  a  different  value  at  different 
times. 

14.  The  value  of  a  figure  depends  upon  its  position  in 
relation  to  other  figures. 

15.  Figures  have  simple  values,  and  local,  or  relative, 

values. 

16.  The  simple  value  of  a  figure  is  the  value  it  expresses 
when  standing  alone. 

17.  The   local,  or   relative,  value   of   a   figure   is  the 
increased  value  it  expresses  by  having  other  figures  placed 
on  its  right. 

For  instance,  if  we  see  the  figure  6  standing  alone, 

thus 6 

we  consider  it  as  six  units,  or  simply  six. 

Place  another  6  to  the  left  of  it ;  thus 66 

The  original  figure  is  still  six  units,  but  the  second 
figure  is  ten  times  6,  or  6  tens. 

If  a  third  6  be  now  placed  still  one  place  further 
to  the  left,  it  is  increased  in  value  ten  times  more, 

thus  making  it  6  hundreds 666 

A  fourth  6  would  be  6  thousands 6666 

A  fifth  6  would  be  6  tens  of  thousands,  or  sixty 

thousand 66666 

A  sixth  6  would  be  6  hundreds  of  thousands  .   .      666666 
A  seventh  6  would  be  6  millions 6666666 


§1 


ARITHMETIC. 


The  entire  line  of  seven  figures  is  read  six  millions  six 
hundred  sixty-six  thousands  six  hundred  sixty-six. 

18.  The  increased  value  of  each  of  these  figures  is  its 
local,  or  relative,  value.     Each  figure  is  ten  times  greater  in 
value  than  the  one  immediately  on  its  right. 

19.  The  cipher  (0)  has  no  value  in  itself,  but  it  is  useful 
in  determining  the  place  of  other  figures.     To  represent  the 
number  four  hundred  five ^  two  digits  only  are  necessary,  one 
to  represent  four  hundred  and  the  other  to  represent  five 
units;  but  if  these  two  digits  are  placed  together,  as  45,  the  4 
(being  in  the  second  place)  will  mean  4  tens.    To  mean  4  hun- 
dreds, the  4  should  have  two  figures  on  its  right,  and  a  cipher 
is  therefore  inserted  in  the  place  usually  given  to  tens,  to  show 
that  the  number  is  composed  of  hundreds  and  units  only,  and 
that  there  are  no  tens.     Four  hundred  five  is  therefore  ex- 
pressed as  405.     If  the  number  were  four  thousand  and  five, 
two  ciphers  would  be  inserted ;  thus,  4005.     If  it  were  four 
hundred  fifty,  it  would  have  the  cipher at  the  right-hand  side 
to  show  that  there  were  no  units,  and  only  hundreds  and  tens; 
thus,  450.    Four  thousand  and  fifty  would  be  expressed  4050. 

20.  In  reading  figures,  it  is  usual  to  point  offihe  number 
into  groups  of  three  figures  each,  beginning  at  the  right- 
hand,  or  units,  column,  a  comma  (,)  being  used  to  point  off 
these  groups. 


Billions. 

Millions. 

Thousands. 

Units. 

S' 

en 

1 

_o 

3 

1 

3 

% 

& 

• 

O 

£ 

a 

en 

| 

i 

1 

1 

i 

S 

2 

"8 

£3 

"o 

^ 

Q 

o 

r* 

rn 

"8 

'3 

Hundreds 

Tens  of  Bi 

Billions. 

Hundreds 

Tens  of  M 

Millions. 

Hundreds 

Tens  of  Tl 

Thousands 

Hundreds 

Tens  of  U 

OT 

'3 

4      3      2,1      9      8,7      G      5,4      3      2 

In  pointing  off  these  figures,  begin  at  the  right-hand  figure 
and  count — units,  tens  of  units,  hundreds  of  units;  the  next 


4  ARITHMETIC.  §  1 

group  of  three  figures  is  thousands;  therefore,  we  insert  a 
comma  (,)  before  beginning  with  them.  Beginning  at  the 
figure  5,  we  say  thousands,  tens  of  thousands,  hundreds  of 
thousands,  and  insert  another  comma.  We  next  read  millions, 
tens  of  millions,  hundreds  of  millions  (insert  another  comma). 
billions,  tens  of  billions,  hundreds  of  billions. 

The  entire  line  of  figures  would  be  read:  Four  hundred 
thirty-two  billions  one  htindred  ninety-eight  millions  seven 
hundred  sixty- five  thousands  four  hundred  thirty-two.  When 
we  thus  reads,  line  of  figures,  it  is  called  numeration,  and  if  the 
numeration  be  changed  back  to  figures,  it  is  called  notation. 

For  instance,  the  writing  of  the  following  figures, 

72,584,623, 

would  be  the  notation,  and  the  numeration  would  be  seventy- 
two  millions  Jive  hundred  eighty-four  thousands  six  hundred 
twenty-three. 

21.  NOTE. — It  is  customary   to  leave   the    "s"   off   the   words 
millions,  thousands,  etc.  in  cases  like  the  above,  both  in  speaking  and 
writing;   hence,  the  above  would  usually   be  expressed   seventy-two 
million  five  hundred  eighty-four  thousand  six  hundred  twenty-three. 

22.  The  four  fundamental  processes  of  arithmetic  are 
addition,  subtraction,  multiplication,  and  division.    They  are 
called  fundamental  processes  because  all  operations  in  arith- 
metic are  based  upon  them. 


ADDITION. 

23.  Addition  is  the  process  of  finding  the  sum  of  two  or 
more  numbers.     The  sign  of  addition  is  -f.      It  is  read  plus, 
and  means  more.     Thus,   5  +  6  is  read  5  plus  6,  and  means 
that  5  and  6  are  to  be  added. 

24.  The  sign  of  equality  is  =.     It  is  read  equals  or  is 
equal  to.     Thus,  5  -f  6  =  11  may  be  read  5  plus  6  equals  11, 
or  5  plus  6  is  equal  to  11. 

25.  Like  numbers  can  be  added;  unlike  numbers  cannot 
be  added.     6  dollars  can  be   added  to   7   dollars,    and  the 
sum  will  be  13  dollars;  but  6  dollars  cannot  be  added  to 
7  feet. 


§1 


ARITHMETIC. 


26.  The  following  table  gives  the  sum  of  any  two 
numbers  from  1  to  12 ;  it  should  be  carefully  committed  to 
memory : 


1  and    1  is    2 

2  and    1  is    3 

3  and    1  is    4 

4  and    1  is    5 

1  and    2  is    3 

2  and    2  is    4 

3  and    2  is    5 

4  and    2  is    6 

1  and    3  is    4 

2  and    3  is    5 

3  and    3  is    6 

4  and    3  is    7 

1  and    4  is    5 

2  and    4  is    6 

Sand    4  is    7 

4  and    4  is    8 

1  and    5  is    6 

2  and    5  is    7 

3  and    5  is    8 

4  and    5  is    9 

1  and    6  is    7 

2  and    6  is    8 

3  and    6  is    9 

4  and    6  is  10 

1  and    7  is    8 

2  and    7  is    9 

3  and    7  is  10 

4  and    7  is  11 

1  and    8  is    9 

2  and    8  is  10 

3  and    8  is  11 

4  and    8  is  12 

1  and    9  is  10 

2  and    9  is  11 

3  and    9  is  12 

4  and    9  is  13 

1  and  10  is  11 

2  and  10  is  12 

3  and  10  is  13 

4  and  10  is  14 

1  and  11  is  12 

2  and  11  is  13 

3  and  11  is  14 

4  and  11  is  15 

1  and  12  is  13 

2  and  12  is  14 

3  and  12  is  15 

4  and  12  is  16 

5  and    1  is    6 

6  and    1  is    7 

7  and    1  is    8 

8  and    1  is    9 

5  and    2  is    7 

6  and    2  is    8 

7  and    2  is    9 

8  and    2  is  10 

5  and    3  is    8 

6  and    3  is    9 

7  and    3  is  10 

8  and    3  is  11 

5  and    4  is    9 

6  and    4  is  10 

7  and    4  is  11 

8  and    4  is  12 

5  and    5  is  10 

6  and    5  is  11 

7  and    5  is  12 

8  and    5  is  13 

5  and    6  is  11 

6  and    6  is  12 

7  and    6  is  13 

8  and    6  is  14 

5  and    7  is  12 

6  and    7  is  13 

7  and    7  is  14 

8  and    7  is  15 

5  and    8  is  13 

6  and    8  is  14 

7  and    8  is  15 

8  and    8  is  16 

5  and    9  is  14 

6  and    9  is  15 

7  and    9  is  16 

8  and    9  is  17 

5  and  10  is  15 

6  and  10  is  16 

7  and  10  is  17 

8  and  10  is  18 

5  and  11  is  16 

6  and  11  is  17 

7  and  11  is  18 

8  and  11  is  19 

5  and  12  is  17 

6  and  12  is  18 

7  and  12  is  19 

8  and  12  is  20 

9  and    1  is  10 

10  and    1  is  11 

11  and    1  is  12 

12  and    1  is  13 

9  and    2  is  11 

10  and    2  is  12 

11  and    2  is  13 

12  and    2  is  14 

9  and    3  is  12 

10  and    3  is  13 

11  and    3  is  14 

12  and    3  is  15 

9  and    4  is  13 

10  and    4  is  14 

11  and    4  is  15 

12  and    4  is  16 

9  and    5  is  14 

10  and    5  is  15 

11  and    5  is  16 

12  and    5  is  17 

9  and    6  is  15 

10  and    6  s  16 

11  and    6  is  17 

12  and    6  is  18 

9  and    7  is  16 

10  and    7  s  17 

11  and    7  is  18 

12  and    7  is  19 

9  and    8  is  17 

10  and    8  s  18 

11  and    8  is  19 

12  and    8  is  20 

9  and    9  is  18 

10  and    9  s  19 

11  and    9  is  20 

12  and    9  is  21 

9  and  10  is  19 

10  and  10  s  20 

11  and  10  is  21 

12  and  10  is  22 

9  and  11  is  20 

10  and  11   s  21 

11  and  11  is  22 

12  and  11  is  23 

9  and  12  is  21 

10  and  12  is  22 

11  and  12  is  23    • 

12  and  12  is  24 

27.  For  addition,  place  the  numbers  to  be  added  directly 
under  one  another,  taking  care  to  place  units  under  units, 
tens  under  tens,  Jiundreds  under  hundreds  ^  and  so  on. 

When  the  numbers  are  thus  written,  the  right-hand  figure 
of  one  number  is  placed  directly  under  the  right-hand  figure 
of  the  one  above  it,  thus  bringing-  units  under  units,  tens 
under  tens,  etc.  Proceed  as  in  the  following  examples: 


•• 


6  ARITHMETIC.  §  1 

28.  EXAMPLE.— What  is  the  sum  of  131,  222,  21,  2,  and  413  ? 

SOLUTION. —  131 

222 

21 

2 

413 

sum  789    Ans. 

EXPLANATION. — After  placing  the  numbers  in  proper 
order,  begin  at  the  bottom  of  the  right-hand,  or  units,  col- 
umn and  add;  thus,  3  +  2  +  1  +  2  +  1  =  9,  the  sum  of  the 
numbers  in  units  column.  Place  the  9  directly  beneath  as 
the  first,  or  units,  figure  in  the  sum. 

The  sum  of  the  numbers  in  the  next,  or  tens,  column 
equals  8  tens,  which  is  the  second,  or  tens,  figure  in  the 
sum. 

The  sum  of  the  numbers  in  the  next,  or  hundreds,  col- 
umn equals  7  hundreds,  which  is  the  third,  or  hundreds, 
figure  in  the  sum. 

The  sum,  or  answer,  is  789. 

29.  EXAMPLE.— What  is  the  sum  of  425,  36,  9,215,  4,  and  907  ? 
SOLUTION.—  425 


sum  10587    Ans. 

EXPLANATION. — The  sum  of  the  numbers  in  the  first,  or 
units,  column  is  27  units;  i.  e.,  2  tens  and  7  units.  Write  27 
as  shown.  The  sum  of  the  numbers  in  the  second,  or  tens, 
column  is  6  tens,  or  60.  Write  60  underneath  27  as  shown. 
The  sum  of  the  numbers  in  the  third,  or  hundreds,  column 
is  15  hundreds,  or  1,500.  Write  1,500  under  the  two  pre- 
ceding results  as  shown.  There  is  only  one  number  in  the 
fourth,  or  thousands,  column,  9,  which  represents  9,000. 


§  1  ARITHMETIC.  7 

Write  9,000  under  the  three  preceding  results.  Adding 
these  four  results,  the  sum  is  10,587,  which  is  the  sum  of  425, 
36,  9,215,  4,  and  907. 

3O.     The  addition  may  also  be  performed  as  follows: 
425 
36 

9215 

4 

907 


sum  10587    Ans. 

EXPLANATION. — The  sum  of  the  numbers  in  units  col- 
umn =  27  units,  or  2  tens  and  7  units.  Write  the  7  units 
as  the  first,  or  right-hand,  figure  in  the  sum.  Reserve  the 
2  tens  and  add  them  to  the  figures  in  tens  column.  The 
sum  of  the  figures  in  the  tens  column  plus  the  2  tens 
reserved  and  carried  from  the  units  column  is  8,  which  is 
written  down  as  the  second  figure  in  the  sum.  There  is 
nothing  to  carry  to  the  next  column,  because  8  is  less 
than  10.  The  sum  of  the  numbers  in  the  next  column  is 
15  hundreds,  or  1  thousand  and  5  hundreds.  Write  down 
the  5  as  the  third,  or  hundreds,  figure  in  the  sum  and  carry 
the  1  to  the  next  column.  1  -j-  9  =  10,  which  is  written 
down  at  the  left  of  the  other  figures. 

The  second  method  saves  space  and  figures,  but  the  first 
is  to  be  preferred  when  adding  a  long  column. 

31.     EXAMPLE.— Add  the  numbers  in  the  column  below  : 
SOLUTION.—  890 

82 

90 
393 
281 

80 
770 

83 
492 

80 
383 

84 
191 


sum  3899  Ans. 
H.  S.  I.—2 


8  ARITHMETIC.  §  1 

EXPLANATION. — The  sum  of  the  digits  in  the  first  column 
equals  19  units,  or  1  ten  and  9  units.  Write  the  9  and 
carry  1  to  the  next  column.  The  sum  of  the  digits  in  the 
second  column  +  the  1  carried  from  the  first  column  is 
109  tens,  or  10  hundreds  and  9  tens.  Write  down  the  9  and 
carry  the  10  to  the  next  column.  The  sum  of  the  digits  in 
this  column  plus  the  10  carried  from  the  second  column  is  38. 

The  entire  sum  is  3,899. 

32.  Rule. — I.     Begin     at   the    right,   add  each  column 
separately,  and  write  the  sum,  if  it  be  only  one  figure,  under 
the  column  added. 

II.  If  the  sum  of  any  column  consists  of  two  or  more  fig- 
ures, put  the  right-hand  figure  of  the  sum  under  that  column 
and  add  the  remaining  figure  or  figures  to  the  next  column. 

33.  Proof. — To  prove  addition,   add  each  column  from 
top  to  bottom.     If  you  obtain  the  same  result  as  by  adding 
from  bottom  to  top,  the  work  is  probably  correct. 


EXAMPLES  FOR  PRACTICE. 

34.  Find  the  sum  of: 

(a)  104  +  203  +  613  +  214. 

(£)  1,875  +  3,143  +  5,826  +  10,832. 

(<r)  4,865  +  2,145  +  8,173  +  40,084. 

(d)  14,204  +  8,173  +  1,065  +  10,042. 

(e)  10,832  +  4,145  +  3,133  +  5,872. 
(/)  214  +  1,231  +  141  +  5,000. 

(g)  123  +  104  +  425  +  126  +  327. 

(h)     6,354  +  2,145  +  2,042  +  1,111  +  3,333. 


Ans.  - 


(a)  1,134. 

(6)  21,676. 

(c)  55,267. 

(d)  33,484. 


CO   6,586. 
(g)  1,105. 

(A)    14,985. 


1.  A  week's  record  of  coal  burned  in  an  engine  room  is  as  fol- 
lows:    Monday,  1,800  pounds;   Tuesday,  1,655  pounds;   Wednesday, 
1,725  pounds;  Thursday,  1,690  pounds;  Friday,  1,648  pounds;  Satur- 
day, 1,020  pounds.     How  much  coal  was  burned  during  the  week  ? 

Ans.  9,538  Ib. 

2.  A  steam  pump,  in  one  hour,  pumps  out  of  a  cistern  4,200  gal- 
lons ;  in  the  next  hour,  5,420  gallons ;  and  in  45  minutes  more  an  addi- 
tional 3,600  gallons,  when  the  cistern  becomes  empty.     How  many 
gallons  were  in  the  cistern  at  first  ?  Ans.  13,220  gal. 


§  1  ARITHMETIC.  9 

3.  What  is  the  total  cost  of  a  steam  plant,  the  several  items  of 
expense  being  as  follows:  steam  engine,  $900;  boiler,  §775;  fittings 
and  connections,  $225 ;  erecting  the  plant.  $125 :  engine  house,  $650  ? 

Ans.  $2.675. 


SUBTRACTION. 

35.  In  arithmetic,  subtraction  is  the  process  of  finding 
how  much  greater  one  number  is  than  another. 

The  greater  of  the  two  numbers  is  called  the  minuend. 

The  smaller  of  the  two  numbers  is  called  the  subtra- 
hend. 

The  number  left  after  subtracting  the  subtrahend  from 
the  minuend  is  called  the  difference,  or  remainder. 

36.  The  sign  of   subtraction   is  — .      It  is  read  minus, 
and  means  less.     Thus,  12  —  7  is  read  12  minus  7,  and  means 
that  7  is  to  be  taken  from  12. 

37.  EXAMPLE.— From  7,568  take  3,425. 
SOLUTION. —  minuend    7568 

subtrahend    3425 


remainder    4143    Ans. 

EXPLANATION. — Begin  at  the  right-hand,  or  units,  column 
and  subtract  in  succession  each  figure  in  the  subtrahend 
from  the  one  directly  above  it  in  the  minuend,  and  write 
the  remainders  below  the  line.  The  result  is  the  entire 
remainder. 

38.  When  there  are  more  figures  in  the  minuend  than 
in  the  subtrahend,  and  when  some  figures  in  the  minuend 
are  less  than  the  figures  directly  under  them  in  the  sub- 
trahend, proceed  as  in  the  following  example: 

EXAMPLE.— From  8,453  take  844. 

SOLUTION. —  minuend    8453 

subtrahend       844 


remainder    7609    Ans. 

EXPLANATION. — Begin  at  the   right-hand,   or  units,  col- 
umn to  subtract.      We  cannot  take  4  from  3,  and  must, 


10  ARITHMETIC.  §  1 

therefore,  borrow  1  from  5  in  tens  column  and  annex  it  to 
the  3  in  units  column.  The  1  ten  =  10  units,  which  added 
to  the  3  in  units  column  ~  13  units.  4  from  13  =  9,  the 
first,  or  units,  figure  in  the  remainder. 

Since  we  borrowed  1  from  the  5,  only  4  remains;  4  from 
4  =  0,  the  second,  or  tens,  figure.  We  cannot  take  8  from  4, 
so  borrow  1  thousand,  or  10  hundreds,  from  8;  10  hundreds 
+  4  hundreds  =  14  hundreds,  and  8  from  14  =  6,  the  third, 
or  hundreds,  figure  in  the  remainder. 

Since  we  borrowed  1  from  8,  only  7  remains,  from  which 
there  is  nothing  to  subtract ;  therefore,  7  is  the  next  figure 
in  the  remainder,  or  answer. 

The  operation  of  borrowing  is  placing  1  before  the  figure 
following  the  one  from  which  it  is  borrowed.  In  the  above 
example  the  1  borrowed  from  5  is  placed  before  3,  making 
it  13,  from  which  we  subtract  4.  The  1  borrowed  from  8  is 
placed  before  4,  making  14,  from  which  8  is  taken. 

39.  EXAMPLE.— Find  the  difference  between  10,000  and  8,763. 
SOLUTION. —  minuend    10000 

subtrahend       8763 

remainder       1237    Ans. 

EXPLANATION. — In  the  above  example  we  borrow  1  from 
the  second  column  and  place  it  before  0,  making  10 ;  3  from 
10  =  7.  In  the  same  way  we  borrow  1  and  place  it  before 
the  next  cipher,  making  10;  but  as  we  have  borrowed  1 
from  this  column  and  have  taken  it  to  the  units  column, 
only  9  remains  from  which  to  subtract  6 ;  6  from  9  =  3. 
For  the  same  reason  we  subtract  7  from  9  and  8  from  9 
for  the  next  two  figures,  and  obtain  a  total  remainder  of 
1,237. 

40.  Rule. — Place  the   subtrahend   (or  smaller]   number 
under  the  minuend  (or  larger]  rtumber,  in   the  same  manner 
as  for  addition,  and  proceed  as  in  Arts.  37,  38,  and  39. 

41.  Proof. —  To  prove  an  example  in  subtraction,  add  the 
subtrahend  and  the  remainder.      The  sum  should  equal  the 
minuend.     If  it   docs   not,   a  mistake  has   been    made,  and 
the  work  should  be  done  over. 


§  1  ARITHMETIC.  11 

Proof  of  the  above  example : 

subtrahend       8763 
remainder       1237 


minuend    10000 


EXAMPLES    FOR    PRACTICE. 
42.     From : 

(a)  94,278  take  62,574. 

(b)  53,714  take  25,824. 


(c)  71,832  take  58,109. 

(d)  20,804  take  10,408. 

(e)  310,465  take  102,141. 

(/)  (81,043  +  1,041)  take  14,831. 

(g)  (20,482  +  18,216)  take  21,214. 

(k)  (2,040  +  1,213  +  542)  take  3,791. 


Ans. 


(a)  31,704. 

(b)  27, 

(c)  13,723. 

(d)  10,396. 

(e)  208,324. 
(/)  67,253. 
(g)  17,484. 
(k)  4. 


1.  A  cistern  is  fed  by  two  pipes  which  supply  1,200  and  2,250  gallons 
per  hour,  respectively,  and  is  being  emptied  by  a  pump  which  delivers 
5,800  gallons  per  hour.     Starting  with  8,000  gallons  in  the  cistern,  how 
much  water  does  it  contain  at  the  end  of  an  hour  ?  Ans.  5,650  gal. 

2.  A  train  in  running  from  New  York  to  Buffalo  travels  38  miles  the 
first  hour,  42  the  second,  39  the  third,  56  the  fourth,  52  the  fifth,  and  48 
the  sixth  hour.     How  many  miles  remain  to  be  traveled  at  the  end  of 
the  sixth  hour,  the  distance  between  the  two  places  being  410  miles  ? 

Ans.  135  mi. 

3.  On  Monday  morning  a  bank  had  on  hand  $2,862.     During  the 
day  81,831  were  deposited  and  §2,172  drawn  out;  on  Tuesday,  $3,126 
were  deposited  and  $1,954  drawn  out.    How  many  dollars  were  on  hand 
Wednesday  morning  ?  Ans.  $3,693. 


MULTIPLICATION. 

43.  To  multiply  a  number  is  to  add  it  to  itself  a  certain 
number  of  times. 

44.  Multiplication  is  the  process  of   multiplying  one 
number  by  another. 

The  number  thus  added  to  itself,  or  the  number  to  be 
multiplied,  is  called  the  multiplicand. 

The  number  which  shows  how  many  times  the  multiplicand 
is  to  be  taken,  or  the  number  by  which  we  multiply,  is  called 
the  multiplier. 

The  result  obtained  by  multiplying  is  called  the  product. 


ARITHMETIC. 


45.     In  the  following  table,  the  product  of  any  two  num- 
bers (neither  of  which  exceeds  12)  may  be  found: 


1  times    1  is      1 

2  times    1  is      2 

3  times    1  is      3 

:   times    2  is      2 

2  times    2  is      4 

3  times    2  is      6 

'.  times    3  is      3 

•2  times    3  is      6 

3  times    3  is      9 

times    4  is      4 

2  times    4  is      8 

3  times    4  is    12 

times    5  is      5 

2  times    5  is    10 

3  times    5  is    15 

:  times    6  is   .  6 

2  times    6  is    12 

3  times    6  is    18 

times    7  is      7 

2  times    7  is    14 

3  times    7  is    21 

.     times    8  is      8 

2  times    8  is    16 

3  times    8  is    24 

times    9  is      9 

2  times    9  is    18 

3  times    9  is    27' 

times  10  is    10 

2  times  10  is    20 

3  times  10  is    30 

1  times  11  is    11 

2  times  11  is    22 

3  times  11  is    33 

1  times  12  is    12 

2  times  12  is    24 

3  times  12  is    36 

4  times    1  is     4 

5  times    1  is      5 

6  times    1  is      6 

4  times    2  is      8 

5  times    2  is    10 

.6  times    2  is    12 

4  times    3  is    12 

5  times    3  is    15 

6  times    3  is    18 

4  times    4  is    16 

5  times    4  is    20 

6  times    4  is    24 

4  times    5  is    20 

5  times    5  is    25 

6  times    5  is    30 

4  times    6  is    24 

5  times    6  is    30 

6  times    6  is    36 

4  times    7  is    28    . 

5  times    7  is    35 

6  times    7  is    42 

4  times    8  is    32 

5  times    8  is    40 

6  times    8  is    48 

4  times    9  is    36 

5  times    9  is    45 

6  times    9  is    54 

4  times  10  is    40 

5  times  10  is    50 

6  times  10  is    60 

4  times  11  is    44 

5  times  11  is    55 

6  times  11  is    66 

4  times  12  is    48 

5  times  12  is    60 

6  times  12  is    72 

7  times    1  is      7 

8  times    1  is      8 

9  times    1  is      9 

7  times    2  is    14 

8  times    2  is    16 

9  times    2  is    18 

7  times    3  is    21 

8  times    3  is    24 

9  times    3  is    27 

7  times    4  is    28 

8  times    4  is    32 

9  times    4  is    36 

7  times    5  is    35 

8  times    5  is    40 

9  times    5  is    45 

7  times    6  is    42 

8  times    6  is    48 

9  times    6  is    54 

7  times    7  is    49 

8  times    7  is    56 

9  times    7   s    63 

7  times    8  is    56 

8  times    8  is    64 

9  times    8  s    72 

7  times    9  is    63 

8  times    9  is    72 

9  times    9   s    81 

7  times  10  is    70 

8  times  10  is    80 

9  times  10  s    90 

7  times  11  is    77 

8  times  11  is    88 

9  times  11    s    99 

7  times  12  is    84 

8  times  12  is    96 

9  times  12  s  108 

10  times    1  is    10 

11  times    1  is    11 

12  times    1  is    12  «.- 

10  times    2  is    20 

11  times    2  is    22 

12  times    2  is    24 

10  times    3  is    30 

11  times    3  is    33 

12  times    3  is    36 

10  times    4  is   40 

11  times    4  is    44 

12  times    4  is    48 

10  times    5  is    50 

11  times    5  is    55 

12  times    Sis    60 

10  times    6  is    60 

11  times    6  is    66 

12  times    6  is    72 

10  times    7  is    70 

11  times    7  is    77 

12  times    7  is    84 

10  times    Sis    80 

11  times    8  is    88 

12  times    8  is    96 

10  times    9  is    90 

11  times    9  is    99 

12  times    9  is  108 

10  times  10  is  100 

11  times  10  is  110 

12  times  10  is  120 

10  times  11  is  110 

11  times  11  is  121 

12  times  11  is  132 

10  times  12  is  120 

11  times  12  is  132 

12  times  12  is  144 

This  table  should  be  carefully  committed  to  memory. 
Since  0  has  no  value,  the  product  of  0  and  any  number  is  0. 


§  1  ARITHMETIC.  13 

46.  The  sign  of  multiplication  is  X  .     It  is  read  times  or 
multiplied  by.     Thus,  9  X  6  is  read  9  times  6,  or  9  multi- 
plied by  6. 

47.  It  matters  not  in  what  order  the  numbers  to  be 
multiplied  together  are  placed.     Thus,   6  X  9  is  the   same 
as  9  X  6. 

48.  To  multiply  a  number  by  one  figure  only : 

EXAMPLE. — Multiply  425  by  5. 
SOLUTION.—  multiplicand       425 

multiplier  5 

product    2125    Ans. 

EXPLANATION. — For  convenience,  the  multiplier  is  gener- 
ally written  under  the  right-hand  figure  of  the  multiplicand. 
On  looking  in  the  multiplication  table,  we  see  that  5  X  5  is  25. 
Multiplying  the  first  figure  at  the  right  of  the  multiplicand, 
or  5,  by  the  multiplier  5,  it  is  seen  that  5x5  units  is  25  units, 
or  2  tens  and  5  units.  Write  the  5  units  in  units  place  in  the 
product  and  reserve  the  2  tens  to  add  to  the  product  of  tens. 
Looking  in  the  multiplication  table  again,  we  see  that  5x2 
is  10.  Multiplying  the  second  figure  of  the  multiplicand  by 
the  multiplier  5,  we  see  that  5  times  2  tens  is  10  tens,  which, 
plus  the  2  tens  reserved,  is  12  tens,  or  1  hundred  and  2  tens. 
Write  the  2  tens  in  tens  place  and  reserve  the  1  hundred  to 
add  to  the  product  of  hundreds.  Again,  we  see  by  the  multi- 
plication table  that  5  X  4  is  20.  Multiplying  the  third,  or 
last,  figure  of  the  multiplicand  by  the  multiplier  5,  we  see 
that  5  times  4  hundreds  is  20  hundreds,  which,  plus  the 
1  hundred  reserved,  is  21  hundreds,  or  2  thousands  and 
1  hundred,  which  we  write  in  thousands  and  hundreds  places, 
respectively. 

Hence,  the  product  is  2,125. 

This  result  is  the  same  as  adding  425  five  times.     Thus, 

425 
425 
425 
425 
425 


sum  2125  Ans. 


U  ARITHMETIC.  §  1 


EXAMPLES  FOB  PRACTICE. 

49.     Find  the  product  of: 


(a)  61,483  X6. 

(b)  12,375  X  5. 

(c)  10,426  X  7. 

(d)  10,835X3. 

(e)  98,376  X  4. 
(/)  10,873X8. 
(£•)  71,543  X  9. 
(X)  218,734  X  2. 


(a)  368,898. 

(£)  61,875. 

(c)  72,982. 

(<f)  32,505. 

(e)  393,504. 

(/)  86,984. 

(£•)  643,887. 

(X)  437,468. 


1.  A  stationary  engine  makes  5,520  revolutions  per  hour.    Running 
9  hours  a  day,  5  days  in  the  week,  and  5  hours  on  Saturday,  how  many 
revolutions  would  it  make  in  4  weeks  ?  Ans.  1,104,000  rev. 

2.  An  engineer  earns  $650  a  year  and  his  average  expenses  are  $548. 
How  much  could  he  save  in  8  years  at  that  rate  ?  Ans.  §816. 

3.  The  connection  between  an  engine  and  boiler  is  made  up  of 
5  lengths  of  pipe,  three  of  which  are  12  feet  long,  one  2  feet  6  inches 
long,  and  one  8  feet  6  inches  long.     If  the  pipe  weighs  9  pounds  per 
foot,  what  is  the  total  weight  of  the  pipe  used  ?  Ans.  423  Ib. 


5O.     To  multiply  a  number  by  two  or  more 

EXAMPLE.— Multiply  475  by  234. 

SOLUTION. —        multiplicand  475 

multiplier  234 

1900 
1425 
950 


product    111150    Ans. 

EXPLANATION. — For  convenience,  the  multiplier  is  gen- 
erally written  under  the  multiplicand,  placing  units  under 
units,  tens  under  tens,  etc. 

We  cannot  multiply  by  234  at  one  operation ;  we  must, 
therefore,  multiply  by  the  parts  and  then  add  the  partial 
products. 

The  parts  by  which  we  are  to  multiply  are  4  units,  3  tens, 
and  2  hundreds.  4  times  475  =  1,900,  the  first  partial 


ARITHMETIC. 


15 


product ;  3  times  475  =  1,425,  the  second  partial  product,  the 
right-hand  figure  of  which  is  zvritten  directly  under  the  fig- 
ure multiplied  by ,  or  3;  2  times  475=  950,  the  third  partial 
product,  the  right-hand  figure  of  which  is  written  directly 
under  the  figure  multiplied  by,  or  2. 

The  sum  of  these  three  partial  products  is  111,150,  which 
is  the  entire  product. 

51.  Rule. — I.      Write  the  multiplier  under  the  multipli- 
cand, so  that  units  are  under  units,  tens  under  tens,  etc. 

II.  Begin  at   the  right  and  multiply  each  figure  of  the 
•multiplicand  by  each  successive  figure  of  the  multiplier,  pla- 
cing the  right-hand  figure  of  each  partial  product  directly 
under  the  figure  used  as  a  multiplier. 

III.  The  sum  of  the  partial  products    will   equal   the 
required  product. 

52.  Proof. — Review  the  work  carefully,  or  multiply  the 
multiplier  by  the  multiplicand ;  if  the  results  agree,  the  work 
is  correct. 


53.  When  there  is  a  cipher  in  the  multiplier,  multiply  by 
it  the  same  as  with  the  other  figures ;  since  the  result  will 
be  zero,  place  a  cipher  under  the  cipher  in  the  multiplier  and 
write  the  next  partial  product  to  the  left  of  it,  as  shown 
below : 


(a) 

0 

X0 


Ans. 


(*) 

2 
X0 


Ans. 


15 
X_0 

0 


Ans. 


708 

X 0 

0    Ans. 


w 

3114 
203 

9342 
62280 

632142  Ans. 


4008 
305 

20040 
120240 

1222440    Ans. 


31264 
1002 

62528 
3126400 

31326528  Ans. 


16 


ARITHMETIC. 


§1 


EXAMPLES  FOR  PRACTICE 

54.     Find  the  product  of: 
(a)    3,842X26. 


(6)     3,716X45. 

(c)  1,817  X  124. 

(d)  675  X  38. 

(e)  1,875x33. 
(/)   4,836x47. 
(g)  5,682x543. 

(k)    3,257  X  246.  Ans.  -| 

(/)    2,875  X  302. 
(j)    17,819  X  1,004. 
(k)    38,674X205. 
(/)    18,304  X  100. 
(m)   7,834  X  10. 
'   (»)    87,543  X  1,000. 
(0)    48,763  X  100. 

1.  If  the  area  of  a  steam-engine  piston  is  113  square  inches,  what  is 
the  total  pressure  upon  it  when  the  steam  pressure  is  85  pounds  per 
square  inch  ?  Ans.  9,605  Ib. 

2.  A  steam  engine  which  indicated  164  horsepower  was  found  to 
consume  4  pounds  of  coal  per  horsepower  per  hour.     Being  replaced  by 
a  new  engine  of  the  same  horsepower,  another  test  was  made,  which 
showed  a  consumption  of  3  pounds  per  horsepower  per  hour.     What 
was  the  saving  of  coal  for  a  year  of  309  days,  if  the  engine's  average 
run  was  14  hours  a  day  ?  Ans.  709,464  Ib. 

3.  Two  steamers  are  7,846  miles  apart  and  are  sailing  towards  each 
other,  one  at  the  rate  of  18  miles  an  hour  and  the  other  at  the  rate  of 
15  miles  an  hour.     How  far  apart  will  they  be  at  the  end  of  205  hours  ? 

Ans.  1,081  mi. 


(a) 

99,892. 

(*) 

167,220. 

(*) 

225,308. 

(d) 

25,650. 

(e) 

61,875. 

(/) 

227,292. 

(g) 

3,085,326. 

(A) 

801,222. 

(0 

868,250. 

(/) 

17,890,276. 

(*) 

7,928,170. 

(/) 

1,830,400. 

(M) 

78,340. 

(») 

87,543,000. 

(") 

4,876,300. 

DIVISION. 

55.  Division  is  the  process  of  finding  how  many  times 
one  number  is  contained  in  another  of  the  same  kind. 

The  number  to  be  divided  is  called  the  dividend. 

The  number  by  which  we  divide  is  called  the  divisor. 

The  number  which  shows  how  many  times  the  divisor  is 
contained  in  the  dividend  is  called  the  quotient. 


56.     The  sign  of  division  is 
54  4-  9  is  read  54  divided  by  9. 


'-.     It  is  read  divided  by. 
Another  way  to  write  54 


§  1  ARITHMETIC.  17 

divided  by  9  is  ^.  Thus,  54  -=-  9  =  0,  or  —  =  6.  The  expres- 
sion —  may  be  read  either  54  divided  by  9  or  54  over  9,  the 

y 

word  "  over  "  implying  that  there  is  a  line  between  the  two 
numbers  indicating  that  the  upper  number  is  to  be  divided 
by  the  lower  one.  In  both  of  these  cases  54  is  the  dividend 
and  9  is  the  divisor. 

Division  is  the  reverse  of  multiplication. 

57.  To  divide  when  the  divisor  consists  of  but  one 
figure,  proceed  as  in  the  following  example: 

EXAMPLE. — What  is  the  quotient  of  875  -r-  7  ? 

divisor  dividend  quotient 

SOLUTION. —  7)875(1  25     Ans. 

7 

17. 

14 

~35 
35 

remainder        0 

EXPLANATION. —  7  is  contained  in  8  hundreds,  1  hundred 
times.  Place  the  1  as  the  first,  or  left-hand,  figure  of 
the  quotient.  Multiply  the  divisor,  7,  by  the  1  hundred  of 
the  quotient,  and  place  the  product,  7  hundreds,  under  the 
8  hundreds  in  the  dividend,  and  subtract.  Beside  the 
remainder,  1,  bring  down  the  7  tens,  making  17  tens; 
17  divided  by  7  =  2  tens.  Write  the  2  as  the  second  figure 
of  the  quo'tient.  Multiply  the  divisor,  7,  by  the  2,  and 
subtract  the  product  from  17.  Beside  the  remainder,  3, 
bring  down  the  5  units  of  the  dividend,  making  35  units. 
7  is  contained  in  35,  5  times,  and  5  is  placed  in  the  quotient. 
5  times  7  =  35,  which,  subtracted  from  35,  under  which  it  is 
placed,  leaves  0.  Therefore,  the  quotient  is  125.  This 
method  is  called  long  division. 

58.  In    short    division,     only    the    divisor,    dividend, 
and  quotient  are  written,  the  operations  being  performed 
mentally. 


18  ARITHMETIC.  §  1 

dividend 

divisor    7  )8'735 
quotient    125    Ans. 

The  mental  operation  is  as  follows:  7  is  contained  in  8, 
1  time  and  1  remainder  ;  1  placed  before  7  makes  17  ;  7  is  con- 
tained in  17,  2  times  and  3  over;  the  3  placed  before  5 
makes  35  ;  7  is  contained  in  35,  5  times.  These  partial 
quotients,  placed  in  order  as  they  are  found,  make  the  entire 
quotient,  125. 

59.  If  the  divisor  consists  of  tzvo  or  more  figures,  pro- 
ceed as  in  the  following  example: 

EXAMPLE.—  Divide  2,702,826  by  63. 

divisor         dividend          quotient 

SOLUTION.—  63)2702826(42902    Ans. 

252 


126 


567 

126 
126 

EXPLANATION.  —  As  63  is  not  contained  in  the  first  two 
figures,  27,  we  must  use  the  first  three  figures,  270.  Now, 
by  trial  we  must  find  how  many  times  63  is  contained  in  270. 
6  is  contained  in  the  first  two  figures  of  270,  4  times.  Place 
the  4  as  the  first,  or  left-hand,  figure  in  the  quotient.  Mul- 
tiply the  divisor,  63,  by  4,  and  subtract  the  product,  252, 
from  270.  The  remainder  is  18,  beside  which  we  write  the 
next  figure  of  the  dividend,  2,  making  182.  Now,  6  is  con- 
tained in  the  first  two  figures  of  182,  3  times,  but  on  multi- 
plying 63  by  3,  we  see  that  the  product,  189,  is  too  great,  so 
we  try  2  as  the  second  figure  of  the  quotient.  Multiplying 
the  divisor,  63,  by  2  and  subtracting  the  product,  126,  from 
182,  the  remainder  is  56,  beside  which  we  bring  down  the 
next  figure  of  the  dividend,  making  568.  6  is  contained  in 
56  about  9  times.  Multiply  the  divisor,  63,  by  9  and  sub- 
tract the  product,  567,  from  568.  The  remainder  is  1,  and 


§  1  ARITHMETIC.  19 

bringing  down  the  next  figure  of  the  dividend,  2,  gives  12. 
As  12  is  smaller  than  63,  we  write  0  in  the  quotient  and 
bring  down  the  next  figure,  6,  making  126.  63  is  contained 
in  126,  2  times,  without  a  remainder.  Therefore,  42,902  is 
the  quotient. 

60.  Rule.— I.      Write  the  divisor  at  the  left  of  the  divi- 
dend, wit /i  (i  line  bet^veen  them. 

II.  Find  how  many  times  the  divisor  is  contained  in  the 
lowest  number  of  the  left-hand  figures  of  the  dividend  that 
will  contain  it,  and  write  the  result  at  the  riglit  for  t lie  first 
figure  of  the  quotient. 

III.  Multiply   the   divisor   by   this   quotient;    write   the 
product  under  the  partial  dividend  used,  and  subtract,  and  to 
the  remainder  annex  the  next  figure  of  the  dividend.     Divide 
as  before,  and  thus  continue  until  all  the  figures  of  the  divi- 
dend have  been  used. 

IV.  If  any  partial  dividend  will  not  contain  the  divisor, 
write  a  cipher  in  the  quotient,  annex  the  next  figure  of  the 
dividend,  and  proceed  as  before. 

V.  If  there  be  at  last  a  remainder,  write  it  after  the  quo- 
tient, with  the  divisor  underneath. 

61.  Proof. — Multiply  t  lie  quotient  by  the  divisor  and  add 
the  remainder,  if  there  be  any,  to  the  product.      The  result 
will  be  the  dividend.     Thus, 

divisor  dividend  quotient 
63)4235(6  7£|     Ans. 
378 

455 
441 

remainder  1  4 

PROOF,  quotient  6  7 

divisor  6  3 

801 

402 


4221 
remainder  1  4 


dividend     4235 


20  ARITHMETIC. 

EXAMPLES    FOR    PRACTICE. 
62.      Divide  the  following: 
(«)      126,498  by  58. 
(6)     3,207,594  by  767. 


(c)  11,408,202  by  234. 

(d)  2,100,315  by  581. 

(e)  969,936  by  4,008. 
(/)  7,481,888  by  1,021. 
(g)  1,525,915  by  5,003. 
(X)  1,646,301  by  381. 


Ans. 


(a)  2,181. 

(b)  4,182. 

(c)  48,753. 

(d)  3,615. 

(e)  242. 
(/)  7,328. 
(g)  305. 
(A)  4,321. 


1.  In  a  mile  there  are  5,280  feet.     How  many  rails  would  it  take  to 
lay  a  double  row  1  mile  long,  each  rail  being  30  feet  long  ? 

Ans.  352  rails. 

2.  How  many  rivets  will  be  required  for  the  longitudinal  seams  of  a 
cylindrical  boiler  20  feet  long,  the  joint  being  double-riveted,  and  the 
rivets  being  spaced  4  inches  apart  ?  Ans.   120  rivets. 

NOTE. — First  find  the  length  of  the  boiler  in  inches. 

3.  It  requires  7,020,000  bricks  to  build  a  large  foundry.    How  many 
teams  will  it  require  to  draw  the  bricks  in  60  days,  if  each  team  draws 
6  loads  per  day  and  1,500  bricks  at  a  load  ?  Ans.  13  teams. 

NOTE.— Find  how  many  loads ' 7,020,000  bricks  make;    then,  how 
many  days  it  will  take  one  team  to  draw  the  bricks. 


CAlTOELATICXtf. 

63.  Cancelation  is  the  process  of  shortening  operations 
in  division  by  casting  out  equal  factors  from  both  dividend 
and  divisor. 

64.  The  factors  of  a  number  are  -those  numbers  which, 
when  multiplied  together,  will  equal  that  number.     Thus, 
5  and  3  are  the  factors  of  15,  since  5  X  3  =  15.      Likewise, 
8  and  7  are  the  factors  of  56,  since  8  X  7  =  56. 

65.  A  prime  number  is  one  which  cannot  be  divided 
by  any  number  except  itself  and  1.     Thus,  2,  3,  11,  29,  etc. 
are  prime  numbers. 

66.  A  prime  factor  is  any  factor  that  is  a  prime  number. 
Any  number  that  is  not  a  prime  is  called  a  composite 

number,  and  may  be  produced  by  multiplying  together  its 


§  1  ARITHMETIC.  21 

prime  factors.     Thus,  60  is  a  composite  number,  and  is  equal 
to  the  product  of  its  prime  factors,  2x2x3x5. 

Numbers  are  said  to  be  prime  to  each  other  when  no  two 
of  them  can  be  divided  by  any  number  except  1 ;  the  numbers 
themselves  may  be  either  prime  or  composite.  Thus,  the 
numbers  3,  5,  and  11  are  prime  to  one  another ;  so  also  are  22, 
25,  and  21 — all  three  of  which  are  composite  numbers. 

67.  Canceling  equal  factors  from  both  dividend  and  divisor 
does  not  change  the  quotient. 

Canceling  a  factor  in  both  dividend  and  divisor  is  the  same 
as  dividing'  them  both  by  the  same  number,  which,  by  the 
principle  of  division,  does  not  change  the  quotient. 

Write  the  numbers  forming  the  dividend  above  the  line, 
and  those  forming  the  divisor  below  it. 

68.  EXAMPLE.— Divide  4  X  45  X  60  by  9  X  24. 

SOLUTION. — Placing  the  dividend  over  the  divisor,  and  canceling, 


Ans. 


EXPLANATION. — The  4  in  the  dividend  and  the  24  in  the 
divisor  are  both  divisible  by  4,  since  4  divided  by  4  equals  1, 
and  24  divided  by  4  equals  6.  Cross  off  the  4  and  write  the  1 
over  it :  also  cross  off  the  24  and  write  the  6  under  it.  Thus. 


60  in  the  dividend  and  6  in  the  divisor  are  divisible  by  6, 
since  60  divided  by  6  equals  10,  and  6  divided  by  6  equals  1. 
Cross  off  the  60  and  write  10  over  it;  also  cross  off  the  6  and 
write  1  under  it.  Thus, 


1  10 

t  X  45  X  99 


Again,  45  in  the  dividend  and  9  in  the  divisor  are  divisible 
by  9,   since  45  divided  by  9  equals  5,  and  9  divided  by  9 


22  ARITHMETIC.  §  1 

equals  1.     Cross  off  the  45  and  write  the  5  over  it;  also  cross 
off  the  9  and  write  the  1  under  it.     Thus, 
l      5      10 


i     ? 

Since  there  are  no  two  remaining  numbers  (one  in  the  divi- 
dend and  one  in  the  divisor)  divisible  by  any  number  except  1, 
without  a  remainder,  it  is  impossible  to  cancel  further. 

Multiply  all  the  uncanceled  numbers  in  the  dividend 
together  and  divide  their  product  by  the  product  of  all  the 
uncanceled  numbers  in  the  divisor.  The  result  will  be 
the  quotient.  The  product  of  all  the  uncanceled  numbers  in 
the  dividend  equals  5  X  1  X  10  =  50;  the  product  of  all  the 
uncanceled  numbers  in  the  divisor  equals  lXl  —  1. 
1  5  10 

B^-xfxfi.mx".*  Ans 

1    p 

It  is  usual  to  omit  the  1's  when  canceling,  instead  of  wri- 
ting them  as  above. 

69.  Rule.  —  I.  Cancel  the  common  factors  from  both  the 
dividend  and  the  divisor. 

II.  Then  divide  the  product  of  the'  remaining  factors  of  the 
dividend  by  the  product  of  the  remaining'  factors  of  the  divisor, 
and  the  result  will  be  the  quotient. 


EXAMPLES  FOB  PRACTICE. 
7O.     Divide : 

(a)     14  X  18X16X40  by  7  X  8  X  6  X  5  X  3. 
(6)      3  X  65  X  50  X  100  X  60  by  30  X  60  X  13  X  10. 

(c)  8  X  4  X  3  X  9  X  11  by  11  X  9  X  4  X  3  X  8. 

(d)  164  X  321  X  6  X  7  X  4  by  82  X  321  x  7. 

(e)  50  X  100  X  200  X  72  by  1,000  X  144  X  100. 
(/)    48  X  63  X  55  X  49  by  7  X  21  x  11  X  48. 
(g)    110  X  150  X  84  x  32  by  11  x  15  X  100  X  64. 

(A)     115  x  120  X  400  X  1,000  by  23  X  1,000  X  60  X  800. 


Ans.  - 


(a)  32. 

(£)  250. 

(0  1- 

(d)  48. 

(e)  5. 
(/)  105. 
U)  42. 
(A)  5. 


ARITHMETIC. 

(PART  2.) 

FRACTIONS. 

REMARK. — If  a  stick  of  wood  is  divided  into,  say,  12  equal 
parts,  one  of  these  parts  is  called  a  twelfth.  If  we  take 
away  5  of  these  equal  parts,  we  shall  have  left  7  parts  or 
7-twelfths.  Since  it  would  be  very  inconvenient  to  spell  out 
the  names  of  the  number  of  parts  into  which  an  object  has 
been  (or  is  supposed  to  have  been)  divided,  mathematicians 
invented,  long  ago,  a  kind  of  a  shorthand  method  of  express- 
ing 7-twelfths,  25-forty-fifths,  etc.,  viz.:  they  wrote  the 
number  of  the  equal  parts  taken  or  considered  above  a 
horizontal  line,  and  called  this  number  the  numerator; 
then  they  wrote  below  the  horizontal  line  the  number  which 
denoted  the  number  of  equal  parts  into  which  the  object  was 
supposed  to  be  divided,  and  called  it  the  denominator. 
Hence,  instead  of  writing  7-twelfths,  25-forty-fifths,  etc., 
they  wrote  T\,  ||,  etc. ;  but  they  read  them  the  same  as  if 
they  had  been  written  the  other  way. 

1.  A  fraction  is  one  or  more  equal  parts  of  a  unit.      One- 
half,  two-thirds,  seven-fifths  are  fractions. 

2.  Two  numbers  are  required  to  express  a  fraction ;  one 
is  called  the  numerator  and  the  other  the  denominator. 

3.  The  numerator  is  placed  above  the  denominator  with 
a  line  between  them,  as  f .      Here  3  is  the  denominator,  and 
shows  into  how  many  equal  parts  the  unit,  or  one,  is  divided. 
The  numerator  2  shows  how  many  of  these  equal  parts  are 

§1 

For  notice  of  copyright,  see  page  immediately  following  the  title  page, 

H.  8.    I.— 3 


24  ARITHMETIC.  §  1 

taken  or  considered.     The  denominator  also  indicates  the 
names  of  the  parts. 

£  is  read  one-half. 

f  is  read  three-fourths. 

£  is  read  three-eighths. 

TST  is  read  five-sixteenths. 

In  the  expression  "f  of  an  apple,"  the  denominator,  4, 
shows  that  the  apple  is  cut  into  4  equal  parts,  and  the 
numerator,  3,  shows  that  three  of  these  parts,  or  fourths,  are 
taken  or  considered. 

If  each  of  the  parts,  or  fourths,  of  the  apple  were  cut  into 
two  equal  pieces,  one  of  these  pieces  would  be  \  of  ^,  or  ^  of 
the  whole  apple,  and  three  of  the  pieces  would  be  •§  of  the 
apple.  Thus,  it  is  evident  that  the  larger  the  denominators, 
the  smaller  the  parts  into  which  anything  is  divided  and 
the  less  the  value  of  the  fraction,  the  numerators  being  the 
same.  Thus,  £  is  less  than  f . 

4.  The  value  of  a  fraction  is  the  numerator  divided  by 
the  denominator,  as  f  =  2,  -f  =  3. 

5.  The  line  between  the  numerator  and  the  denominator 
means  divided  by,  or  -=-. 

f  is  equivalent  to  3  -r-  4. 
f  is  equivalent  to  5  -r-  8. 

6.  The  numerator  and  the  denominator  of  a  fraction  are 
called  the  terms  of  a  fraction. 

7.  The  value  of  a  fraction  whose  numerator  and  denomi- 
nator are  equal  is  1. 

f ,  or  four-fourths  =  1. 

•f,  or  eight-eighths  =  1. 

-££,  or  sixty-four  sixty-fourths  ==  1. 

8.  A  proper  fraction  is  a  fraction  whose  numerator  is 
less  than  its  denominator.     Its  value  is  less  than  1,  as  f ,  f ,  T1T. 

9.  An  Improper  fraction  is  a  fraction  whose  numerator 
is  equal  to  or  is  greater  than  the  denominator.     Its  value  is 
1  or  more  than  1,  as  f ,  •£,  £f. 

10.  A  mixed  number  is  a  whole  number  and  a  fraction 
united.     4f  is  a  mixed  number  and  is  equivalent  to  4  +  f . 
It  is  read  four  and  two-thirds. 


§  1  ARITHMETIC.  25 

REDUCTION    OF    FRACTIONS. 

11.  Reduction  of  fractions  is  the  process  of  changing 
their  forms  without  changing  their  value. 

12.  A  fraction  is  reduced  to  higher  terms  by  multiplying 
both  terms  of  the  fraction  by  the  same  number.     Thus,  £  is 
reduced  to  -f  by  multiplying  both  terms  by  2. 

3  X  2  _6 

4  X  2  ~  8' 

The  value  is  not  changed,  for  f  —  |. 

13.  A  fraction  is  reduced  to  loiver  terms  by  dividing  both 
terms  by  the  same  number.     Thus,  T8¥  is  reduced  to  f.by  divi- 
ding both  terms  by  2. 

_8_-f-  2  _  4 

10  -T-  2  ~~  5* 

14.  A  fraction  is  reduced  to  lowest  terms  when  its  numer- 
ator and  denominator  cannot  be  divided  by  the  same  number, 
as  |,  f,  H- 

15.  To  reduce  a  -whole  number  or  a  mixed  number 
to  an  Improper  fraction : 

EXAMPLE  1. — How  many  fourths  in  5  ? 

SOLUTION. — Since  there  are  4  fourths  in  1  (|  =  1),  in  5  there  will  be 
5x4  fourths,  or  20  fourths;  i.  e.,  5  X  f  =  *£••     Ans. 

EXAMPLE  2. — Reduce  8f  to  an  improper  fraction. 
SOLUTION.—    8  x  £  =  V-     ¥  +  *  =  ¥•     Ans- 

16.  Rule. — Multiply  the  whole  number  by  the  denomi- 
nator of  the  fraction,  add  the  numerator  to  the  product,  and 
place  the  denominator  under  the  result. 


EXAMPLES  FOR  PRACTICE. 

1*7.    Reduce  to  improper  fractions: 
(a)     4$. 


(c)  10A-  Ans-   < 

(d)  37$. 
(')     50*. 


W- 


(e)     if*. 


26  ARITHMETIC.  §  1 

18.  To  reduce  an  improper  fraction  to  a  -whole  or 
a  mixed  number  : 

EXAMPLE.—  Reduce  ^  to  a  mixed  number. 

SOLUTION.—  4  is  contained  in  21,  5  times  and  1  remaining;  as  this 
is  also  divided  by  4,  its  value  is  £.  Therefore,  5  +  i,  or  5^,  is  the  num- 
ber. Ans. 

19.  Rule.  —  Divide  the  numerator   by  the  denominator, 
the  quotient  will  be  the  whole  number;    the  remainder,   if 
there  be  any,  will  be  the  numerator  of  the  fractional  part,  of 
which  the  denominator  is  the  same  as  the  denominator  of  the 
improper  fraction. 

EXAMPLES   FOB  PRACTICE. 

20.  Reduce  to  whole  or  mixed  numbers: 

(a)    i*s.  f(«)     24*. 


n 
(a)    if*.  IS'J(*0    49*. 

(')     If-  (*)     4. 

(/)  W-  U/)  5. 

21.  A  common  denominator  of  two  or  more  fractions 
is  a  number  which  will  contain  all  the  denominators  of  the 
fractions  without  a  remainder.    The  least  common  denom- 
inator is  the  least  number  that  will  contain  all  the  denom- 
inators of  the  fractions  without  a  remainder. 

22.  To  find  the  least  common  denominator  : 

EXAMPLE.  —  Find  the  least  common  denominator  of  £,  -|,  £,  and  T\. 
SOLUTION.—  We  first  place  the  denominators  in  a  row,  separated  by 
commas. 

2  )  4,     3,     9.  16 

2  )  2,  3,  9,  8 

3  )  1,  3,  9.  4 
8  )  1.  1,  3,  4 

4  )  1,  1.  1.  4 


1,     1,     1,     1 
2x2x3x3x4  =  144,  the  least  common  denominator.     Ans. 

EXPLANATION.  —  Divide  each  of  them  by  some  prime  num- 
ber which  will  divide  at  least  two  of  them  without  a  remain- 
der (if  possible),  bringing  down  those  denominators  to  the 


§  1  ARITHMETIC.  27 

row  below  which  will  not  contain  the  divisor  without  a 
remainder.  Dividing  each  of  the  numbers  by  2,  the  second 
row  becomes  2,  3,  9,  8,  since  2  will  not  divide  3  and  9  with- 
out a  remainder.  Dividing  again  by  2,  the  result  is  1,  3,  9,  4. 
Dividing  the  third  row  by  3,  the  result  is  1,  1,  3,  4.  So 
continue  until  the  last  row  contains  only  1's.  The  product 
of  all  the  divisors,  or  2x2x3x3x4=  144,  is  the  least 
common  denominator. 

23.  EXAMPLE. — Find  the  least  common  denominator  of  f,  •£$,  -£s. 
SOLUTION.—  3  )  9,  12,  18 

3)3,     4,     6 

2)1,    4,    2 

2)1,     2,     1 

1,     1,     1 

3x3x2x2  =  36.     Ans. 

24.  To   reduce  two  or  more  fractions  to  fractions 
having  a  common  denominator  : 

EXAMPLE. — Reduce  $,  f,  and  |  to  fractions  having  a  common 
denominator. 

SOLUTION. — The  common  denominator  is  a  number  which  will  con- 
tain 3,  4,  and  2.  The  least  common  denominator  is  12,  because  it  is 
the  smallest  number  which  can  be  divided  by  3,  4,  and  2  without  a 
remainder.  t  ==  A,  f  =  A,  *  =  A 

Reducing  f,  3  is  contained  in  12,  4  times.  By  multiplying  both 
numerator  and  denominator  of  £  by  4,  we  find 

3  X  4  =  12'     In  the  same  way  we  find  *  =  A  and  ^  =  A- 

25.  Rule. — Divide  the  common  denominator  by  the  denomi- 
nator of  the  given  fraction  and  multiply  both  terms  of  the 
fraction  by  the  quotient. 


EXAMPLES  FOR  PRACTICE. 

26.     Reduce  to  fractions  having  a  common  denominator: 

(a)     t,  *,  *•  f(«)       I,    I,    1- 

(*)      T3*-  t-  A-  (*)      A.  tt,  A- 

(0     *,  A.  H-  An,  I  W  H.  A.  II- 

1. 1,  H-  1  (<0  «.  tt.  H- 

(')     A.  A,  -A-  to  if,  A.  it- 

CO 


28  ARITHMETIC.  §  1 

ADDITION  OF  FRACTIONS. 

27.  Fractions  cannot  be  added  unless  they  have  a  common 
denominator.     We  cannot  add  £  to  £•  as  they   now  stand, 
since  the  denominators  represent  parts    of   different    sizes. 
Fourths  cannot  be  added  to  eighths. 

The  fractions  should  be  reduced  to  the  least  common 
denominator,  or  the  least  number  which  will  contain  all  the 
denominators. 

28.  EXAMPLE.  —  Find  the  sum  of  £,  f  ,  and  $•. 

SOLUTION.  —  The  least  common  denominator,  or  the  least  number 
which  will  contain  all  the  denominators,  is  8. 

|  =  f,   f  =  f,    and  f  =  |. 

EXPLANATION.  —  As  the  denominator  tells  or  indicates  the 
names  of  the  parts,  the  numerators  only  are  added  to  obtain 
the  total  number  of  parts  indicated  by  the  denominator. 

t+i+«=*±f±«=i5=ii.  An,       : 

29.  EXAMPLE  1.—  What  is  the  sum  of  12f,  14f  ,  and  7T<V  ? 
SOLUTION.  —  The  least  common  denominator  in  this  case  is  16. 


sum  33  +  f  I  =  33  +  \\\  =  34^-    Ans. 

The  sum  of  the  fractions  =  fj  or  1{£,  which  added  to  the  sum  of  the 
whole  numbers  =  34^. 

EXAMPLE  2.—  What  is  the  sum  of  17,  13T»ff,  fa  and  3±  ? 
SOLUTION.—  The  least  common  denominator  is  32.      13T\  =  13s6j, 
8i  =  3,V  17 


sum    33  1|    Ans. 

3O.  Rule.  —  I.  Reduce  the  given  fractions  to  fractions 
having  the  least  common  denominator  and  write  the  sum  of 
the  numerators  over  the  common  denominator. 


§1 


ARITHMETIC. 


29 


II.  When  there  are  mixed  numbers  and  whole  numbers, 
add  tlie  fractions  first,  and  if 'their  sum  is  an  improper  frac- 
tion, reduce  it  to  a  mixed  number  and  add  the  whole  number 
with  the  other  whole  numbers, 


31, 


EXAMPLES  FOB  PRACTICE. 

Find  the  sum  of: 
(a) 
W) 
(c) 

(d) 


00 
te) 

(4) 


t.  A.  f • 

I.  T5F-  «• 
i,  t,   A- 

f ,  H.  if- 
if,  A.  If- 
If.  H.  if 

TV  A,  if- 

f ,  it.  f  • 


Ans. 


W 


Hi- 

in- 
if 

IA- 
i. 


1.  The    weights    of    a    number    of    castings    were    412f    pounds, 
2704  pounds,  1,020  pounds,  75J  pounds,  and  68|  pounds.     What  was 
their  total  weight  ?  Ans.  1,847  Ib. 

2.  Four  bolts  are  required,  2f,  1|,  2T5^,  and  l^f  inches  long.     How 
long  a  piece  of  iron  will  be  required  to  cut  them  from,  allowing  f  of  an 
inch  altogether  for  cutting  off  and  finishing  the  ends  ?         Ans.  9^  in. 


SUBTRACTION    OF    FRACTIONS. 

32.     Fractions  cannot  be  subtracted  without  first  reducing 
them  to  a  common  denominator. 
EXAMPLE.— Subtract  f  from  f|. 
SOLUTION.— The  common  denominator  is  16. 
18- 


=  A-    H-A  = 


16 


=  Ans. 


Ans. 


33.  EXAMPLE.—  From  7  take  f. 

SOLUTION.—    l  =  f;   therefore,  7=6  +  f  =  6|;   6f  —  |  =  6|. 

34.  EXAMPLE.—  What  is  the  difference  between  17T9^  and  9£|  ? 
SOLUTION.  —  The  common  denominator  of  the  fractions  is  32.     17T9j 


minuend 
subtrahend 

difference 


9f 


Ans. 


30  ARITHMETIC.  §  1 

35.     EXAMPLE.—  From  9£  take  4TV 

SOLUTION.  —  The  common  denominator  of  the  fractions  is  16. 
9i  =  9TV 

minuend    9T\  or  8f  £ 
subtrahend    4T7¥       4T7^ 

difference    4#       4JJ    Ans. 

EXPLANATION.  —  As  the  fraction  in  the  subtrahend  is 
greater  than  the  fraction  in  the  minuend,  it  cannot  be  sub- 
tracted ;  therefore,  borrow  1,  or  -ff  ,  from  the  9  in  the  minuend 
and  add  it  to  the  T\  ;  T4g  +  {£  =  f  £.  ^  from  f  |  =  {£.  Since  1 
was  borrowed  from  9,  8  remains;  4  from  8  =  4;  4  -f-  \\ 


36.  EXAMPLE.—  From  9  take  8TV 

SOLUTION.  —  minuend    9      or  8|f 

subtrahend    &fo       8& 
difference      \\         \\    Ans. 

EXPLANATION.  —  As  there  is  no  fraction  in  the  minuend 
from  which  to  take  the  fraction  in  the  subtrahend,  borrow  1, 
or  -ff  ,  from  9.  T3T  from  4J-  =  -{--§-.  Since  1  was  borrowed 
from  9,  only  8  is  left.  8  from  8  =  0. 

37.  Rule.  —  I.     Reduce  the  fractions  to  fractions  having 
a  common  denominator.     Subtract  one  numerator  from  the 
other  and  place  the  remainder  over  the  common  denominator. 

II.  When  there  are  mixed  numbers,  subtract  the  fractions 
and  whole  numbers  separately,  and  place  the  remainders  side 
by  side. 

III.  When  the  fraction  in  the  subtrahend  is  greater  than 
the  fraction  in  the  minuend,  borrow  1  from  the  whole  number 
in   the  minuend  and  add  it  to  the  fraction   in   the  minuend, 
from  which  subtract  the  fraction  in  the  subtrahend. 

IV.  When    the  mimiend  is  a  whole  number,  borroiv  1  ; 
reduce  it  to  a  fraction  whose  denominator  is  the  same  as  the 
denominator  of  the  fraction  in  the  subtrahend,  and  place  it 
over  that  fraction  for  subtraction. 


§1 


ARITHMETIC. 


31 


EXAMPLES    FOR    PRACTICE. 


38.      Subtract: 


(a) 


to 

(d) 

w 


Ans. 


A- 

u- 
A- 
±. 

17*. 
14*. 
24f. 


^  from  \\. 

TT  from  |f. 

54ff  from  ^. 

^f  from  4§- 

|f  from  ££. 
(f)     13±  from  30f 
(£•)    12£  from  27. 
(h)     5£  from  30. 

1.  An  engineer  found  that  he  had  on  hand  48^  gallons  of  cylinder 
oil.     During  the  following  week  he  used  f  of  a  gallon  each  day  for 
three  days,  *  of  a  gallon  on  the  fourth  day,  if  of  a  gallon  on  the  fifth 
day,  and  i  of  a  gallon  on  the  sixth  day.     How  much  oil  remained  at 
the  end  of  the  week  ?  Ans.  43[f  gal. 

2.  The  main  line  shaft  of  a  manufacturing  plant  is  run  by  an  engine 
and  waterwheel.     A  test  of  the  plant  showed  that  the  engine  was 
capable  of  developing  251|  H.  P.  (horsepower),  and  the  waterwheel, 
under  full  gate,  67£  H.  P.     It  was  also  found  that  the  machinery  con- 
sumed 210|£  H.  P.,  and  the  friction  of  the  shafting  and  belting  was 
32£  H.  P.     How  much  power  remained  unused  ?  Ans.  76f  f  $  H.  P. 


MUI/TIPIjICATIOX    OF    FRACTIONS. 

39.  In  multiplication  of  fractions  it  is  not  necessary  to 
reduce  the  fractions  to  fractions  having  a  common  denominator. 

Multiplying  the  numerator  or  dividing  the  denominator 
multiplies  the  fraction. 

EXAMPLE.  —  Multiply  f  by  4. 


SOLUTION.  — 


f  x  4  = 


'  ' 


=       =  3.     Ans. 


Or, 


=     =  3.     Ans. 


4O.     The  word  "  of  "  in  multiplication  of  fractions  means 
the  same  as  X ,  or  times.     Thus, 

f  of  4  =  |X4  =  3. 
£  of  ft  =  %  X  ft  =  Tf-g-. 
EXAMPLE.— Multiply  f  by  2. 

o   v>.  o 

SOLUTION. —  Sxl^^''     =:f  =  t-     Ans. 

o 

Or,  2  X  f  =  I      9  =  f.     Ans. 

O  -T-  6 


32  ARITHMETIC.  §  1 

41.  EXAMPLE.— What  is  the  product  of  T\  and  |  ? 

4-  v  7 
SOLUTION.-  A  X  f  =^  *  g-  =  i¥s  =  A-     Ans- 

4v7  7 

Or,  by  cancelation,  #T^~8  =  ?^8  =  &'     AnS< 

4 

42.  EXAMPLE.— What  is  |  of  |  of  if  ? 

SOLUTION.-  8  X  ^  X  ^  =  8^2  =  *'     Ans' 

2 

43.  EXAMPLE.— What  is  the  product  of  9f  and  5f  ? 

SOLUTION.—  9|  =  \9- ;  5|  =  -4/-. 

¥X¥  =  ^*f-  =  4f*  =  54B.     Ans. 

44.  EXAMPLE.— Multiply  15  J  by  3. 

SOLUTION.—  15|  15| 

3     or          3 
475  45  +  _2i_  _  45  +  2|  =  47|.     Ans. 

45.  Rule. — I.     Divide  the  product  of  the  numerators  by 
the  product  of  tJie  denominators.     All  factors  common  to  tJie 
numerators  and  denominators  should  first  be  cast  out  by  can- 
celation. 

II.  To  multiply  one   mixed  number   by   another,    reduce 
them  both  to  improper  fractions. 

III.  To  multiply  a  mixed  number  by  a  ^cvhole  number,  first 
multiply  the  fractional  part  by  the  multiplier,   and  if  the 
product  is  an  improper  fraction,  reduce  it  to  a  mixed  num- 
ber and  add  the  whole  number  part  to  the  product  of  the  mul- 
tiplier and  the  whole  number. 


EXAMPLES  FOR  PRACTICE. 

46.      Find  the  product  of: 

(«)      7XTV  f(a)  1TV 

(*>  .  14  X  A.  (6)  4|. 

(0      11  X  A-  (O  **• 

(*)     tf  X  4.  (d)  2tf. 

W      H  X  7.  Ans-  \  (e}  7A- 

(/)     l?if  X  7.  I   (/)  125. 

(£)    m  X  32.  (g)  15. 

(*)      tf  X  14.  (A)  7i. 


§  1  ARITHMETIC.  33 

1.  A  single  belt  can  transmit  107f  horsepower,  but  as  it  is  desired  to 
use  more  power,  a  double  belt  of  the  same  width  is  substituted  for  it. 
Supposing  the  double  belt  to  be  capable  of  transmitting  ^P-  as  much 
power  as  the  single  belt,  how  many  horsepower  can  be  used  after  the 
change  ?  Ans.  153£f  H.  P. 

2.  What  is  the  weight  of  2f  miles  of  copper  wire  weighing  5|  pounds 
per  100  feet  ?    There  are  5,280  feet  in  a  mile.  Ans.  796 JJ  Ib. 

3.  The  grate  of  a  steam  boiler  contains  20 £  square  feet.     If  the 
boiler  burns  8  fa  pounds  of  coal  an  hour  per  square  foot  of  grate  area 
and  can  evaporate  7-J  pounds  of  water  an  hour  per  pound  of  coal 
burned,  how  many  pounds  of  water  are  evaporated  by   the  boiler 
in  1  hour  ?  Ans.  1,276^  Ib. 


DIVISION    OF    FRACTIONS. 

47.  In  division  of  fractions  it  is  not  necessary  to  reduce 
the  fractions  to  fractions  having  a  common  denominator. 

48.  Dividing  the  numerator  or  multiplying  the  denomi- 
nator divides  the  fraction. 

EXAMPLE  1.—  Divide  f  by  3. 

SOLUTION.  —  When  dividing  the  numerator,  we  have 

|-3  =  |^3  =  f  =  i.  Ans. 
When  multiplying  the  denominator,  we  have 

|H-3  =  ^x3  =  ^r  =  i.  Ans. 
EXAMPLE  2.—  Divide  •&  by  2. 

O 

SOLUTION.—  ^^2  =  iQX2  =  A<     AnS* 

EXAMPLE  3.  —  Divide  -J-f  by  7. 

14  —  7 
SOLUTION.—        $|  -*-  7  =  ^  '      =^  =  ^.    Ans. 

49.  To  invert  a  fraction  is  to  turn  it  upside  down;  that 
is,  make  the  numerator  and  denominator  change  places. 

Invert  f  and  it  becomes  -f  . 

50.  EXAMPLE.  —  Divide  T9^  by  T\. 

SOLUTION.  —  1.  The  fraction  TS6  is  contained  in  T9ff,  3  times,  for 
the  denominators  are  the  same,  and  one  numerator  is  contained  in 
the  other  3  times.  2.  If  we  now  invert  the  divisor  -fa  and  mul- 
tiply, the  solution  is 


This  brings  the  same  quotient  as  in  the  first  case. 


34  ARITHMETIC.  §  1 

51.      EXAMPLE.—  Divide  f  by  J. 

SOLUTION.  —  We  cannot  divide  $  by  £,  as  in  the  first  case  above,  for 
the  denominators  are  not  the  same  ;  therefore,  we  must  solve  as  in  the 
second  case. 


2 

52.     EXAMPLE  1.—  Divide  5  by  |£. 
SOLUTION.  —    j£  inverted  becomes  ^f. 


EXAMPLE  2.  —  How  many  times  is  3f  contained  in  7^  ? 
SOLUTION.—  3f  =  Y  ;  V<r  =  W- 

-^-  inverted  equals  T4?. 

119       4  _H9X*_119          . 
16  X  15  ~  J0  X  15  ~  60  ~  lff> 

4 

53.  Rule.  —  Invert  the  divisor  and  proceed  as  in  multipli- 
cation. 

54.  We  have  learned  that  a  line   placed  between  two 
numbers  indicates  that  the  number  above  the  line  is  to  be 
divided  by  the  number  below  it.     Thus,  -1-/-  shows  that  18  is 
to  be  divided  by  3.     This  is  also  true  if  a  fraction  or  a  frac- 
tional expression  be  placed  above  or  below  a  line. 

9  3  y  7 

-  means  that  9  is  to  be  divided  by  f  ;  means  that 

16 
3  X  7  is  to  be  divided  by  the  value  of        .    . 

|  is  the  same  as  £  -H  f  . 

It  will  be  noticed  that  there  is  a  heavy  line  between  the  9 
and  the  f  .  This  is  necessary,  since  otherwise  there  would 
be  nothing  to  show  as  to  whether  9  was  to  be  divided  by  f 
or  f  was  to  be  divided  by  8.  Whenever  a  heavy  line  is  used, 
as  shown  here,  it  indicates  that  all  above  the  line  is  to  be 
divided  by  all  below  it. 


§  1  ARITHMETIC.  35 

55.  Whenever  an  expression  like  one  of  the  three  fol- 
lowing ones  is  obtained,  it  may  always  be  simplified  by 
transposing  the  denominator  from  above  to  below  the  line,  or 
from  beloiv  to  above,  as  the  case  may  be,  taking  care,  how- 
ever, to  indicate  that  the  denominator  when  so  transferred 
is  a  multiplier. 

so  « 

1-     |  =  g-^-j  =  A  =  TIT  J    for,     regarding    the    fraction 

above  the  heavy  line  as  the  numerator  of  a  fraction  whose 
denominator  is  9,  ^         = -,  as  before. 

y  x  4     9x4 

*—  =  12.     The  proof  is  the  same  as  in  the  first 


—  =  £«. ;    for,  regarding  f  as  the  numerator 


5  X  9          5 

of  a  fraction    whose  denominator  is  £ ,  |-         = ;  and 

if  X  9       o  X  9 

5X45X4  4 

=  ^TTTT  =  f  4>  as  above. 


X9X4 


4 

This  principle  may  be  used  to  great  advantage  in  cases  like 

i  X  310  X  14  X  72      _ 

L-7?. ^     — .     Reducing  the  mixed  numbers  to  frac- 

4:1)  X  4^-  X  o-g- 

.      ,  i  X  310  X  H  X  72      __ 

tions,  the  expression  becomes  = — — — -=-j^ .     Now  trans- 
ferring the  denominators  of   the  fractions   and  canceling, 

J0        3        0  3 

1  x  310  x  27  X  72  x  2  x  6  _  1  X  m  X  ?T  X  n  X  %  X  0 
40  X  9  X  31  X  4  X  12  ^0  x  ^  X  n  X  £  X  ft 

I  2 

.=  ¥  =  13*.  ?  O 

Greater  exactness  in  results  can  usually  be  obtained  by 
using  this  principle  than  can  be  obtained  by  reducing  the 
fractions  to  decimals.  The  principle,  however,  should  not 
be  employed  if  a  sign  of  addition  or  subtraction  occurs  either 
above  or  below  the  dividing  line. 


ARITHMETIC. 


EXAMPLES  FOB  PRACTICE. 
56.     Divide: 

(a)    15by6f  (a) 

(£)    SObyf 


(r)     172  by  f 

W)    ttbyl* 
(<?)     io*  by  14f. 


Cr) 


40. 
215. 


00 
(/) 

Cr) 


(A)    J#  by  721. 

1.  A  f-inch  boiler  plate  containing  24  square  feet  of  surface  weighs 
362-j4^  pounds.     What  is  its  weight  per  square  foot  ?  Ans.  15^  Ib. 

2.  A  certain  boiler  has  927£  square  feet  of  heating  surface,  which  is 
equal  to  35  times  the  area  of  the  grate.     What  is  the  area  of  the  grate 
in  square  feet  ?  Ans.  26^  sq.  ft. 

3.  If  the  distance  around  the  rim  of  a  locomotive  driving  wheel  is 
ISy1^  feet,  how  many  revolutions  will  the  wheel  make  in  traveling 

Ans.  522     rev. 


ARITHMETIC. 

(PART  3.) 

DECIMALS. 

1.  Decimals  are  tenth  fractions ;  that  is,  the  parts  of  a 
unit  are  expressed  on  the  scale  of  ten,  as  tenths,  hundredths, 
thousandths,  etc. 

2.  The  denominator,  which  is  always  10,  100,  1,000,  etc., 
is  not  expressed,  as  it  would  be  in  fractions,  by  writing  it  under 
the  numerator  with  a  line  between  them,  as  T3¥,  y^-g-,  T7V?r>  DUt 
is   expressed    by   placing   a   period  (.),  which    is  called   a 
decimal  point,  to  the  left  of  the  figures  of  the  numerator, 
to  indicate  that  the  figures  on  the  right  form  the  numerator 
of  a  fraction  whose  denominator  is  ten,  one  Jnmdred,  one 
thousand,  etc. 

3.  The  reading  of  a  decimal  number  depends  upon  the 
number  of  decimal  places  in  it,  i.  e.,the  number  of  figures 
to  the  right  of  the  decimal  point. 

One  decimal  place  expresses  tenths. 
Two  decimal  places  express  hundredths. 
Three  decimal  places  express  thousandths. 
Four  decimal  places  express  ten-thousandths. 
Five  decimal  places  express  hundred-thousandths. 
Six  decimal  places  express  millionths. 
§1 

For  notice  of  copyright,  see  page  immediately  following  the  title  page. 


38  ARITHMETIC.  §  1 

Thus: 

.3  =       ^       =3  tenths. 

.03          =      yfo      =3  hundredths. 
.003        =     T^O-     =  3  thousandths. 
.0003      =    Tolnro    —  3  ten-thousandths. 
.00003    =  ToifVoir  —  3  hundred-thousandths. 
.000003  =  nnrioTro  =  3  millionths. 

We  see  in  the  above  that  the  number  of  decimal  places  in  a 
decimal  equals  the  number  of  ciphers  to  the  right  of  the  figure  1 
in  the  denominator  of  its  equivalent  fraction.  This  fact  kept 
in  mind  will  be  of  much  assistance  in  reading  and  writing 
decimals. 

Whatever  may  be  written  to  the  left  of  a  decimal  point  is 
a  whole  number.  The  decimal  point  affects  only  the  figures 
to  its  right. 

When  a  whole  number  and  decimal  are  written  together, 
the  expression  is  a  mixed  number.  Thus,  8.12  and  17.25  are 
mixed  numbers. 

The  relation  of  decimals  and  whole  numbers  to  each  other 
is  clearly  shown  by  the  following  table  : 


a 
o 

45 

BJ 

| 

I 

# 
•*j 

rt 

•4-5 

ifi 

m 

£ 

CO 

o 

millions. 

hundreds  of 

O 

os 
1 

thousands. 

hundreds. 

to 

R 

0 

units. 

c 
*o 

OH 

E 
'o 

'O 

09 

V 

hundredths. 

thousandths. 

ten-thousand 

hundred-thoi 

(A 

c 
.2 

'£ 

1 

I 

hundred-mill 

7 

6 

5 

4 

3 

2 

1 

a 

3 

4 

5 

6 

f 

8 

9 

The  figures  to  the  /*//  of  the  decimal  point  represent 
whole  numbers  ;  those  to  the  right  are  decimals. 

In  both  decimals  and  whole  numbers,  the  units  place  is 
made  the  starting  point  of  notation  and  numeration.  The 
decimals  decrease  on  the  scale  of  ten  to  the  right,  and  the 
whole  numbers  increase  on  the  scale  of  ten  to  the  left.  The 
first  figure  to  the  left  of  units  is  tens,  and  the  _/&•*/  figure  to 
the  right  of  units  is  tenths.  The  jrawdT  figure  tQ  the  /<?//  of 


§  1  ARITHMETIC.  39 

units  is  hundreds,  and  the  second  figure  to  the  right  is  Jiun- 
dredths.  The  third  figure  to  the  left  is  thousands,  and  the 
third  to  the  right  is  thousandths,  and  so  on,  the  whole 
numbers  on  the  left  and  the  decimals  on  the  right.  The 
figures  equally  distant  from  units  place  correspond  in  name. 
The  decimals  have  the  ending  ths,  which  distinguishes  them 
from  whole  numbers.  The  following  is  the  numeration  of 
the  number  in  the  above  table:  nine  hundred  eighty-seven 
million  six  hundred  fifty-four  thousand  three  hundred 
twenty-one  and  twenty-three  million  four  hundred  fifty-six 
thousand  seven  hundred  eighty-nine  hundred-millionths. 

The  decimals  increase  to  the  left  on  the  scale  of  ten,  the 
same  as  whole  numbers  ;  for,  beginning  at,  say,  ^-thousandths 
in  the  table,  the  next  figure  to  the  left  is  hundredths,  which 
is  ten  times  as  great,  and  the  next  tenths,  or  ten  times  the 
hundredtJis,  and  so  on  through  both  decimals  and  whole 
numbers. 

4.  Annexing  or  taking  away  a  cipher  at  the  right  of  a 
decimal  does  not  affect  its  value. 

-  .5  is  T57  ;  .50  is  -ffo,  but  TV  =  yW  :  therefore,  .5  =  .50. 

5.  Inserting  a  cipher  between  a  decimal  and  the  decimal 
point  divides  the  decimal  by  10. 


6.     Taking  away  a  cipher  from  the  left  of  a  decimal  multi- 
plies the  decimal  by  10. 

•06  =  -;          X10  =       =  .  5. 


ADDITION    OP    DECIMALS. 

7.  The  only  respect  in  which  addition  of  decimals  differs 
from  addition  of  whole  numbers  is  in  the  placing  of  the 
numbers  to  be  added. 

Whole  numbers  begin  at  units  and  increase  on  the  scale  of  10, 
to  the  left.  Decimals  decrease  on  the  scale  of  10,  to  the  right. 
Whole  numbers  are  to  the  left  of  the  decimal  point,  and 
decimals  are  to  the  right  of  it.  In  whole  numbers  the  right- 
hand  side  of  a  column  of  figures  to  be  added  must  be  in  line, 
H.  s.  I.— 4 


40                                    ARITHMETIC.  §  1 

and  in  decimals  the  left-hand  side  must  be   in  line,  which 
brings  the  decimal  points  directly  under  one  another. 

whole  numbers                                decimals  mixed  numbers 

342                                  .342  342.032 

4234                                  .4234  4234.5 

26                                  .26  26.6782 

3                                  .03  3.06 


sum    4605    Ans.         sum     1.0554    Ans.       sum    4606.2702    Ans. 

8.     EXAMPLE.— What   is  the   sum   of  242,    .36,    118.725,    1.005,    6, 
and  100.1  ? 

SOLUTION.—  242. 

.36 

118.725 
1.005 
6. 
100.1 


sum    468.190    Ans. 

9.  Rule. — Place  the  numbers  to  be  added  so  that  the  deci- 
mal points  will  be  directly  under  one  another.  A  dd  as  in 
whole  numbers,  and  place  the  decimal  point  in  the  sum  directly 
under  the  decimal  points  above. 


EXAMPLES   FOB   PBACTICE. 

1O.     Find  the  sum  of: 
(a)      .2143,  .105,  2.3042,  and  1.1417. 
(6)      783.5,  21.473,  .2101,  and  .7816. 

(c)  21.781,  138.72,  41.8738,  .72,  and  1.413. 

(d)  .3724,  104.15,  21.417,  and  100.042. 

0?)      200.172,  14.105,  12.1465,  .705,  and  7.2.        Ans' 
(/)     1,427.16,  .244,  .32,  .032,  and  10.0041. 
(/)    2,473.1,  41.65,  .7243,  104.067,  and  21.073. 
(X)     4,107.2,  .00375,  21.716,  410.072,  and  .0345. 


(a)  3.7652. 

(b)  805.9647. 

(c)  204.5078. 

(d)  225.9814. 

(e)  234.3285. 
(/)  1,437.7601. 
(g)  2,640.6143. 
(A)  4,539.02625. 


1.  The  estimated  weights  of  the  parts  of  a  return-tubular  boiler  were 
as  follows:     shell,  3,626  pounds;  tubes,  3,564.5  pounds;  manhole  cover, 
ring,   and   yoke,   270.34  pounds;    stays,  etc.,   1,089.4  pounds;     steam 
nozzles,  236.07   pounds;    handhole  covers  and  yokes,  120.25  pounds; 
feedpipe,  34.75  pounds;  boiler  supports,  350.6  pounds.     What  was  the 
total  estimated  weight  of  the  boiler  1  Ans.  9,291.91  Ib. 

2.  A  bill  for  engine-room  supplies  had  the  following  items:     1  waste 
can,  $8.30;    20  feet  of  4-inch  belting,  $11.20;    1   pipe  wrench,  $1.65; 
12  pounds  of  waste,  $0.84;  5  gallons  of  cylinder  oil,  $8.75;  20  gallons  of 
machine  oil,  $24.     How  much  did  the  bill  amount  to  1          Ans.  $54.74. 


§  1  ARITHMETIC.  41 

SUBTRACTION    OF    DECIMALS. 

11.  For  the  same  reason  as  in  addition  of  decimals,  the 
left-hand  figures  of  decimal  numbers  are  placed  in  line  and 
the  decimal  points  under  each  other. 

EXAMPLE.— Subtract  .132  from  .3063. 

SOLUTION.—  minuend    .3063 

subtrahend    .132 


difference    .1743    Ans. 

12.     EXAMPLE. — What  is  the  difference  between  7.895  and  .725  ? 

SOLUTION. —  minuend    7.8  9  5 

subtrahend      .725 


difference    7.1  70,  or  7.17    Ans. 

13.     EXAMPLE.— Subtract  .625  from  11. 

SOLUTION.—  minuend    1  1.0  0  0 

subtrahend         .625 


difference    1  0.3  7  5    Ans. 

14.  Rule. — Place  the  subtrahend  under  the  minuend,  so 
that  the  decimal  points  will  be  directly  under  each  other. 
Subtract  as  in  wJiole  numbers,  and  place  the  decimal  point  in 
the  remainder  directly  under  the  decimal  points  above. 

When  the  figures  in  the  decimal  part  of  the  subtrahend 
extend  beyond  those  in  the  minuend,  place  ciphers  in  the 
minuend  above  them  and  subtract  as  before. 


EXAMPLES  FOR  PRACTICE. 
15.     From; 


(a)  407.385  take  235.0004. 

(b)  22.718  take  1.7042. 

(c)  1,368.17  take  13.6817. 

(d)  70.00017  take  7.000017. 

(e)  630.630  take  .6304. 
(/)  421.73  take  217.162. 
(g)  1.000014  take  .00001. 
(h)  .783652  take  .542314. 


(a)  172.3846. 

(b)  21.0138. 

(c)  1,354.4883. 

(d)  63.000153. 
(€)  629.9996. 
(/)  204.568. 
(g)  1.000004. 
(h)  .241338. 


42  ARITHMETIC.  §  1 

1.  If   the   temperature  of  steam   at  5  pounds  pressure  is  227.964 
degrees  and  at  100  pounds   pressure   is  337.874  degrees,  how  many 
degrees  hotter  is  the  steam  at  the  higher  pressure  ? 

Ans.  109.91  degrees. 

2.  The  outside  diameter  of  2|-inch  wrought-iron  pipe  is  2.87  inches 
and  the  inside  diameter  is  2.46  inches.     How  thick  is  the  pipe  ? 

Ans.   .41  -H  2  =  .205  in. 

3.  In  a  cistern  that  will  hold  326.5   barrels  of  water  there  are 
178.625  barrels  ?    How  much  does  it  lack  of  being  full  ? 

Ans.   147.875  bbl. 

4.  A  wrought-iron  rod  is  2.53  inches  in  diameter.     What  must  be 
the  thickness  of  metal  turned  off,  so  that  the  rod  will  be  2.495  inches  in 
diameter?  Ans.  .035  -=-  2  =  .0175  in. 


MULTIPLICATION    OF    DECIMALS. 

16.  In  multiplication  of  decimals,  we  do  not  place  the 
decimal  points  directly  under  each  other  as  in  addition  and 
subtraction.  We  pay  no  attention  for  the  time  being  to  the 
decimal  points.  Place  the  multiplier  under  the  multiplicand, 
so  that  the  right-hand  figure  of  the  one  is  under  the  riglit- 
hand  figure  of  the  other,  and  proceed  exactly  as  in  multi- 
plication of  whole  numbers.  After  multiplying,  count  t/ic 
number  of  decimal  places  in  both  multiplicand  and  multiplier , 
and  point  off  the  same  number  in  the  product. 

EXAMPLE.— Multiply  .825  by  13. 

SOLUTION.—    multiplicand        .825 
multiplier  1  3 

2475 
825 


product    10.725    Ans. 

In  this  example  there  are  3  decimal  places  in  the  multi- 
plicand and  none  in  the  multiplier;  therefore,  3  decimal 
places  are  pointed  off  in  the  product. 

17.     EXAMPLE.— What  is  the  product  of  426  and  the  decimal  .005  ? 

SOLUTION.—       multiplicand       426 
multiplier      .005 

product    2.1  3  0,  or  2.13    Ans. 


§  1  ARITHMETIC.  43 

In  this  example  there  are  3  decimal  places  in  the  multi- 
plier and  none  in  the  multiplicand;  therefore,  3  decimal 
places  are  pointed  off  in  the  product. 

18.  It  is  not  necessary  to  multiply  by  the  ciphers  on  the 
left  of  a  decimal;  they  merely  determine  the  number  of  deci- 
mal places.     Ciphers  to  the  right  of   a  decimal  should  be 
omitted,  as  they  only  make  more  figures  to  deal  with  and 
do  not  change  the  value. 

19.  EXAMPLE.— Multiply  1.205  by  1.15. 
SOLUTION. — 

multiplicand          1.2  0  5 
multiplier  1.1  5 

6025 
1205 
1205 


product    1.38575    Ans. 

In  this  example  there  are  3  decimal  places  in  the  multi- 
plicand and  2  in  the  multiplier;  therefore,  3  -f-  2,  or  5,  deci- 
mal places  must  be  pointed  off  in  the  product. 

20.  EXAMPLE.— Multiply  .232  by  .001. 

SOLUTION. — 

multiplicand  .232 

multiplier  .001 

product    .000232    Ans. 

In  this  example  we  multiply  the  multiplicand  by  the  digit 
in  the  multiplier,  which  gives  232  for  the  product;  but  since 
there  are  3  decimal  places  each  in  the  multiplier  and  multi- 
plicand, we  must  prefix  3  ciphers  to  the  232  to  make  3  -f-  3, 
or  6,  decimal  places  in  the  product. 

21.  Rule. — Place  tlie  multiplier  under  the  multiplicand, 
disregarding  the  position  of  the  decimal  points.     Multiply  as 
in  -whole  numbers,  and  in  tlie  product  point  off  as  many  deci- 
mal places  as  there  are  decimal  places  in  both  multiplier  and 
multiplicand,  prefixing  ciphers  if  necessary. 


44  ARITHMETIC.  §  1 

EXAMPLES  FOR  PRACTICE. 
22.     Find  the  product  of: 


'(a)  .000492x4.1418. 

(b)  4,003.2  X  1.2. 

(c)  78.6531  X  1-03. 

(d)  .3685  x  -042. 

(e)  178,352  X  .01. 
(/)  .00045  X  .0045. 
(g)  .714  X  .00002. 
(X)  .00004  X  -008. 


(a)  .0020377656. 

(b)  4,803.84. 

(c)  81.012693. 

(d)  .015477. 

(e)  1,783.52.' 
(/)  .000002025. 
(g)  .00001428. 
(k)  .00000032. 


1.  The  stroke  of  an  engine  was  found  by   measurement  to  be 
2.987  feet.     How  many  feet  will  the  crosshead  pass  over  in  600  revolu- 
tions ?  Ans.  3,584.4  ft. 

2.  If  a  steam  pump  delivers  2.39  gallons  of  water  per  stroke  and 
runs  at  51  strokes  a  minute,  how  many  gallons  of  water  would  it  pump 
in  58£  minutes  ?  Ans.  7,130.565  gal. 

3.  Wishing  to  obtain  the  weight  of  a  connecting-rod  from  a  drawing, 
it  was  calculated  that  the  rod  contained  294.8  cubic  inches  of  wrought 
iron,  63.5  cubic  inches  of  brass,  and  10.4  cubic  inches  of  Babbitt.    Assu- 
ming the  weight  of  wrought  iron  to  be  .278  pound  per  cubic  inch,  of 
brass  .303  pound,  and  of  Babbitt  .264  pound,  what  was  the  weight  of 
the  rod?  Ans.  103.94  Ib. 


DIVISION  OF  DECIMALS. 

23.  In  division  of  decimals  we  pay  no  attention  to  the 
decimal  point  until  after  the  division  is  performed.  The 
number  of  decimal  places  in  the  dividend  must  equal  (or  be 
made  to  equal  by  annexing  ciphers]  the  number  of  decimal 
places  in  the  divisor.  Divide  exactly  as  in  whole  numbers. 
Subtract  the  number  of  decimal  places  in  the  divisor  from 
the  number  of  decimal  places  in  the  dividend,  and  point  off 
as  many  decimal  places  in  the  quotient  as  are  indicated  by 
the  remainder. 

EXAMPLE.— Divide  .625  by  25. 

divisor  dividend  quotient 

SOLUTION. —  25).625(.025    Ans. 

50 


125 

125 


remainder 


§  1  ARITHMETIC.  45 

In  this  example  there  are  no  decimal  places  in  the  divisor 
and  3  decimal  places  in  the  dividend;  therefore,  there 
are  3  minus  0,  or  3,  decimal  places  in  the  quotient.  One 
cipher  has  to  be  prefixed  to  the  25  to  make  the  3  deci- 
mal places. 

24:.     EXAMPLE.— Divide  6.035  by  .05. 

divisor    dividend    quotient 

SOLUTION. —  .05)6.035(120.7    Ans. 

5 

10 

12 

35 
35 

remainder      0 

In  this  example  we  divide  by  5,  as  if  the  cipher  were  not 
before  it.  There  is  1  more  decimal  place  in  the  dividend 
than  in  the  divisor;  therefore,  1  decimal  place  is  pointed 
off  in  the  quotient. 

25.  EXAMPLE.— Divide  .125  by  .005. 

divisor    dividend    quotient 

SOLUTION.—  .005). 125(25    Ans. 

10 

~lj5 
25 

remainder      0 

In  this  example  there  are  the  same  number  of  decimal 
places  in  the  dividend  as  in  the  divisor;  therefore,  the  quo- 
tient has  no  decimal  places  and  is  a  whole  number. 

26.  EXAMPLE.— Divide  326  by  .25. 

divisor    dividend    quotient 

SOLUTION.—  .2  5  )  3  2  6.0  0  ( 1  3  0  4    Ans. 

25  , 

76 
75 


100 
100 


remainder 


46  ARITHMETIC.  §  1 

In  this  problem  two  ciphers  were  annexed  to  the  dividend, 
to  make  the  number  of  decimal  places  equal  to  the  number 
in  the  divisor.  The  quotient  is  a  whole  number. 

27.  EXAMPLE.— Divide  .0025  by  1.25. 

divisor     dividend      quotient 

SOLUTION. —  1.2  5  )  .0  0  2  5  0  ( .0  0  2    Ans. 

250 

remainder       0 

EXPLANATION. — In  this  example  we  are  to  divide  .0025  by 
1.25.  Consider  the  dividend  as  a  whole  number,  i.  e. ,  as  25 
(disregarding  the  two  ciphers  at  its  left,  for  the  present) ; 
also,  consider  the  divisor  as  a  whole  number,  i.  e.,  as  125. 
It  is  clearly  evident  that  the  dividend,  25,  will  not  contain 
the  divisor,  125;  we  must,  therefore,  annex  one  cipher  to 
the  25,  thus  making  the  dividend  250.  125  is  contained 
twice  in  250,  so  we  place  the  figure  2  in  the  quotient.  In 
pointing  off  the  decimal  places  in  the  quotient,  it  must  be 
remembered  that  there  were  only  4  decimal  places  in  the 
dividend ;  but  one  cipher  was  annexed,  thereby  making 
4+1,  or  5,  decimal  places.  Since  there  are  5  decimal 
places  in  the  dividend  and  2  decimal  places  in  the  divisor, 
we  must  point  off  5  —  2,  or  3,  decimal  places  in  the  quotient. 
In  order  to  point  off  3  decimal  places,  two  ciphers  must  be 
prefixed  to  the  figure  2,  thereby  making  .002  the  quotient. 
It  is  not  necessary  to  consider  the  ciphers  at  the  left  of  a 
decimal  when  dividing,  except  when  determining  the  posi- 
tion of  the  decimal  point  in  the  quotient. 

28.  Rule.— I.    Place  the  divisor  to  the  left  of  the  dividend 
and  proceed  as  in  division  of  whole  numbers  ;  in  the  quotient, 
point  off  as  many  decimal  places  as  the  number  of  decimal 
places  in  the  dividend  exceeds  the  number  of  decimal  places 
in  the  divisor,  prefixing  ciphers  to  the  quotient,  if  necessary. 

II.  If  in  dividing  one  whole  number  by  another  there  be 
a  remainder,  the  remainder  can  be  placed  over  the  divisor  as 
a  fractional  part  of  the  quotient ;  but  it  is  generally  better  to 
annex  ciphers  to  the  remainder  and  continue  dividing  until 
there  are  three  or  four  decimal  places  in  the  quotient,  and 


§  1  ARITHMETIC.  47 

then  if  there  still  be  a  remainder,  terminate  the  quotient  by 
t/ie  plus  sign  (-(-),  which  shows  that  it  can  be  carried  farther. 

29.     EXAMPLE.— What  is  the  quotient  of  199  divided  by  15  ? 
divisor  dividend  quotient 

SOLUTION.—  15)199(13  + T\    Ans. 

1  5 

45 
remainder        4 

Or,     1  5  )  1  9  9.0  0  0  (  1  3.2  6  6+     Ans. 
15 


45 

~~40 
30 

100 
90 

Foo 


remainder        1  0 

13^  =  13.266+ 
A  =  .266+ 

It  very  frequently  happens,  as  in  the  above  example,  that 
the  division  will  never  terminate.  In  such  cases,  decide  to 
how  many  decimal  places  the  division  is  to  be  carried  and 
carry  the  work  one  place  farther.  If  the  last  figure  of  the 
quotient  thus  obtained  is  5  or  a  greater  number,  increase 
the  preceding  figure  by  1,  and  write  after  it  the  minus  sign 
(  —  ),  thus  indicating  that  the  quotient  is  not  quite  as  large 
as  indicated;  if  the  figure  thus  obtained  is  less  than  5, 
write  the  plus  sign  (+)  after  the  quotient,  thus  indicating 
that  the  number  is  slightly  greater  than  as  indicated.  In 
the  last  example,  had  it  been  desired  to  obtain  the  answer 
correct  to  four  decimal  places,  the  work  would  have  been 
carried  to  five  places,  obtaining  13.26666,  and  the  answer 
would  have  been  given  as  13.2667  —  .  This  remark  applies 
to  any  other  calculation  involving  decimals,  when  it  is 


48  ARITHMETIC.  §  1 

desired  to  omit  some  of  the  figures  in  the  decimal.  Thus, 
if  it  is  desired  to  retain  three  decimal  places  in  the  number 
.2471253,  it  would  be  expressed  as  .247+;  if  it  was  desired 
to  retain  five  decimal  places,  it  would  be  expressed  as  .24713  —  . 
Both  the  -f-  and  —  signs  are  frequently  omitted ;  they  are 
seldom  used  outside  of  arithmetic,  except  in  exact  calcula- 
tions, when  it  is  desired  to  call  particular  attention  to  the 
fact  that  the  result  obtained  is  not  quite  exact. 


EXAMPLES  FOR  PRACTICE. 
3O.     Divide: 

(a)     101.6688  by  2.36.  f  (a)     43.08. 

(£)      187. 12264  by  123.107. 

(c)  .08  by  .008. 

(d)  .0003  by  3. 75.  .          , 
(<?)      .0144  by  .024. 


(/)     .00375  by  1.25. 
(g)    .004  by  400. 
(A)     .4  by  .008. 


(6)  1.52. 

(c)  10. 

(d)  .00008. 

(e)  .6. 


(/)    -003. 
(g)    .00001. 
(A)     50. 


1.  In  a  steam-engine  test  of  an  hour's  duration,  the  horsepower 
developed  was  found  to  be  as  follows  at  10-minute  intervals  :    25.73, 
25.64,  26.13,  25.08,  24.20,  26.7,  26.34.      What  was  the  average  horse- 
power 1  Ans.  25.6886—,  average. 

NOTE. — Add  the  different  horsepowers  together  and  divide  by  the 
number  of  tests,  or  7. 

2.  There  are  31.5  gallons  in  a  barrel.     How  many  barrels  are  there 
in  2,787.75  gallons?  Ans.  88.5  bbl. 

3.  A  carload  of  18.75  tons  of  coal  cost  $60.75.     How  much  was  it 
worth  per  ton  1  Ans.  $3.24  per  ton. 

4.  A  keg  of  Ty  X  If"  boiler  rivets  weighs  100  pounds  and  con- 
tains 595  rivets.     What  is  the  weight  of  one  of  the  rivets  ? 

Ans.  .168+  Ib. 


TO  REDUCE  A  FRACTION  TO  A  DECIMAL. 

31.     EXAMPLE  1. —    4  equals  what  decimal  ? 
SOLUTION.—  4  )  3.0  0 

.7  5,  or  f  =  .75    Ans. 


§1 


ARITHMETIC. 


49 


EXAMPLE  2.— What  decimal  is  equivalent  to 


SOLUTION.  - 


8  )  7.000  (  .875 
64 


60 
56 


or  |  =  .875.     Ans. 


40 
40 


32.  Rule. — Annex  ciphers  to  the  numerator  and  divide 
by  the  denominator.  Point  off  as  many  decimal  places  in  the 
quotient  as  there  are  ciphers  annexed. 


EXAMPLES   FOR   PRACTICE. 

33.     Reduce  the  following  common  fractions  to  decimals: 


(') 


f 

fi- 


to     A- 
(/)   f 


Ans.  - 


(d) 
(*) 
(d) 
(e) 
(/) 

(f) 
(/&) 


.46875. 

.875. 

.65625. 

.796875. 

.16. 

.625. 

.05. 

.004. 


34.     To  reduce  inches  to  decimal  parts  of  a  foot : 

EXAMPLE. — What  decimal  part  of  a  foot  is  9  inches  ? 

SOLUTION. — Since  there  are  12  inches  in  1  foot,  1  inch  is  ^  of  a  foot 
and  9  inches  is  9  X  TJ>  or  T9*  °f  a  foot.  This  reduced  to  a  decimal  by 
the  above  rule  shows  what  decimal  part  of  a  foot  9  inches  is. 

1  2)  9.00  (.7  5  of  afoot.    Ans. 

84 

~60 
60 


35.    Rule. — I.     To  reduce  inches  to  a  decimal  part  of  a 
foot,  divide  the  number  of  inches  by  12. 


50  ARITHMETIC.  §  1 

II.  Should  the  resulting  decimal  be  an  unending  one,  and 
it  is  desired  to  terminate  the  division  at  some  point,  say  tJic 
fourth  decimal  place,  carry  the  division  one  place  farther, 
and  if  the  fifth  figure  is  5  or  greater,  increase  the  fourtJi 
figure  by  1,  omitting  the  signs  -\-  and  — . 


EXAMPLES  FOR  PRACTICE. 

36.      Reduce  to  the  decimal  part  of  a  foot: 

(a)  3  in.  C  (a)  .25  ft. 

(d)  44  in.  (£)  .375  ft. 

(c)  5  in.  Ans.  \  (c)  .4167  ft. 

(d)  6|  in.  (d)  .5521  ft. 

(e)  11  in.  [(*)  .9167ft. 

1.  The  lengths  of  belting  required  to  connect  three  countershafts 
with  the  main  line  shaft  were  found  with  the  tape  measure  to  be  27  feet 
4  inches,  23  feet  8  inches,  and  38  feet  6  inches.     How  many  feet  of 
belting  were  necessary  ?  Ans.  89.5ft. 

2.  The  stroke  of  an  engine  is  14  inches.     What  is  the  length  of  the 
crank  in  feet  measured  from  center  of  shaft  to  center  of  crankpin, 
knowing  that  length  of  crank  is  one-half  the  stroke  ?      Ans.  .5833+  ft. 

3.  A  steam  pipe  fitted  with  an  expansion  joint  was  found  to  expand 
1.668  inches  when  steam  was  admitted  into  it.     How  much   was  its 
expansion  in  decimal  parts  of  a  foot  ?  Ans.  .139  ft. 


TO  REDUCE  A  DECIMAL  TO  A  FRACTION. 

37.  EXAMPLE  1. — Reduce  .125  to  a  fraction. 
SOLUTION.—    .125  =  ^flfr  =  ^  =  $.     Ans. 

EXAMPLE  2.— Reduce  .875  to  a  fraction. 

SOLUTION.—    .875  =  T8^ff  =  ££  _  |.     Ans. 

38.  Rule. —  Under  the  figures  of  the  decimal,  place  1  with 
as  many  ciphers  at  its  right  as  there  are  decimal  places  in  the 
decimal,  and  reduce  the  resulting  fraction  to  its  lowest  terms 
by  dividing  both  numerator  and  denominator  by  the  same 
number. 


§1 


ARITHMETIC. 


51 


EXAMPLES  FOR  PRACTICE. 

30.     Reduce  the  following  to  common  fractions: 


(a)  .375. 

(6)  .625. 

(c)  .3125. 

(d)  .04. 

(e)  .06. 
</)  -75. 
(£•)  .15625. 
(>&)  .875. 


Ans. 


(«)  f- 

(6)  f. 

(0  A- 

(<0  A- 

W  A- 

(/)  t- 


40.  To  express  a  decimal  approximately  as  a  frac- 
tion having  a  given  denominator  : 

EXAMPLE  1.— Express  .5827  in  64ths. 

37  2928 
SOLUTION.—    .5827  X  fj  =      ^      ,  say  |f. 

Hence,  .5827  =  f  J,  nearly.     Ans. 
EXAMPLE  2.— Express  .3917  in  12ths. 
SOLUTION.-    .3917  x  H  =  ^^-  Sa7  A- 
Hence,  .3917  =  &,  nearly.     Ans. 

41.  Rule.— Reduce   1   to   a  fraction   having   the  given 
denominator.     Multiply  the  given   decimal  by  the  fraction 
so  obtained,  and  the  result  will  be  the  fraction  required. 


EXAMPLES  FOR  PRACTICE. 


4:2.     Express: 


(«) 

.625  in  8ths. 

(a)  f. 

(*) 

.3125  in  16ths. 

(b)  ^g. 

w 

.15625  in  32ds. 
Ans 

(?)  A- 

(d} 

.77  in  64ths. 

(</)  !».. 

(') 

.81  in  48ths. 

W  !!• 

(/) 

.923  in  96ths. 

(/)  ft- 

43.     The  sign  for  dollars  is  $.     It  is  read  dollars.     $25  is 
read  £5  dollars. 


52  ARITHMETIC.         .  §1 

Since  there  are  100  cents  in  a  dollar,  1  cent  is  1  one- 
hundredth  of  a  dollar;  the  first  two  figures  of  a  decimal  part 
of  a  dollar  represent  cents.  Since  a  mill  is  TJF  of  a  cent,  or 
TTnnF  of  a  dollar,  the  third  figure  represents  mills. 

Thus,  $25.16  is  read  twenty-jive  dollars  and  sixteen  cents  ; 
$25.168  is  read  twenty-jive  dollars  sixteen  cents  and  eight 
mills.  

SIGNS  OF  AGGREGATION. 

44.  The  vinculum ,  parenthesis  (  ),  brackets  [  ], 

and  brace  {  }  are  called  symbols  of  aggregation,  and  are 
used  to  include  numbers  which  are  to  be  considered  together ; 
thus,  13  X  8  —  3,  or  13  X  (8  —  3),  shows  the  3  is  to  be  taken 
from  8  before  multiplying  by  13. 

13  X  8  —  3  =  13  X  5  =  65.     Ans. 
13  X  (8  -  3)  =  13  X  5  =  65.     Ans. 

When  the  vinculum  or  parenthesis  is  not  used,  we  have 
13x8  —  3=104-3  =  101.     Ans. 

45.  In  any  series  of  numbers  connected  by  the  signs  +, 
— ,  X,  and  -T-,  the  operations  indicated  by  the  signs  must  be 
performed  in  order  from  left  to  right,  except  that  no  addition 
or  subtraction  may  be  performed  if  a  sign  of  multiplication  or 
division  follows  the  number  on  the  right  of  a  sign  of  addition 
or  subtraction  until  the  indicated  multiplication  or  division 
has  been  performed.     In  all  cases  the  sign  of  multiplication 
takes  the   precedence,  the  reason  being  that  when   two  or 
more  numbers  or  expressions   are   connected  by  the  sign  of 
multiplication,  the  numbers  thus  connected  are  regarded  as 
factors  of  the  product  indicated,  and  not  as  separate  numbers. 

EXAMPLE.— What  is  the  value  of  4  x  24  —  8  +  17  ? 
SOLUTION. — Performing  the  operations  in  order  from  left  to  right, 
4  X  24  =  96 ;  96  -  8  =  88 ;  88  +  17  =  105.     Ans. 

46.  EXAMPLE. — What  is  the  value  of  the  following  expression: 
1,296  H-  12  +  160  -  22  x  8$  ? 

SOLUTION.—  1,296  -*-  12  =  108  ;  108  +  160  =  268;  here  we  cannot 
subtract  22  from  268  because  the  sign  of  multiplication  follows  22; 
hence,  multiplying  22  by  3£,  we  get  77,  and  268  —  77  =  191.  Ans. 


§  1  ARITHMETIC.  53 

Had  the  above  expression  been  written  1,296  -f- 12  +  160 
—  22  X  3£  -7-  7  +  25,  it  would  have  been  necessary  to  have 
divided  22  X  3^  by  7  before  subtracting,  and  the  final  result 
would  have  been  22  X  3fc  =  77 ;  77  -=-  7  =  11 ;  268  -11  =  257 ; 
257  +  25  —  282.  Ans.  In  other  words,  it  is  necessary  to  per- 
form all  the  multiplication  or  division  included  between 
the  signs  -f-  and  — ,  or  —  and  4-,  before  adding  or  subtracting. 
Also,  had  the  expression  been  written  1,296  -f-  12  4-  160  —  24^ 
-r-  7  X  34-  4-  25,  it  would  have  been  necessary  to  have  multi- 
plied 3£  by  7  before  dividing  24£,  since  the  sign  of  multiplica- 
tion takes  the  precedence,  and  the  final  result  would  have 
been  3$  X  7  =  24£;  24£  -i-  24.^  =  1 ;  268  -  1  =  267 ;  267  +  25 
=  292.  Ans. 

It  likewise  follows  that  if  a  succession  of  multiplication 
and  division  signs  occurs,  the  indicated  operations  must  not 
be  performed  in  order,  from  left  to  right — the  multiplica- 
tion must  be  performed  first.  Thus,  24x3-^-4x2-^9x5 
=  -\.  Ans.  In  order  to  obtain  the  same  result  that  would 
be  obtained  by  performing  the  indicated  operations  in  order, 
from  left  to  right,  symbols  of  aggregation  must  be  used. 
Thus,  by  using  two  vinculums,  the  last  expression  becomes 
24  X  3~T~4  X  2-7-9  X  5  =  20,  the  same  result  that  would  be 
obtained  by  performing  the  indicated  operations  in  order, 
from  left  to  right. 


EXAMPLES   FOR   PRACTICE. 

4:7.     Find  the  values  of  the  following  expressions: 


(a)  (8  +  5  -  1)  -T-  4. 

(b)  5  X  24  -  32. 
(<:)  5  X  24  -  15. 

(d)  144-5X24. 

(e)  (1,691  —  540  +  559)  -4-  3  X  57. 
(/)  2,080  +  120  —  80  X  4  —  1,670. 
(g)  (90  +  60  H-  25)  X  5  -  29. 

(K)  90  +  60  -5-  25  X  5. 


(a)  3. 

(*)  88. 

(c)  8. 

(d)  24. 

(e)  10. 
(/)  210. 
(/•)  1- 
(A)  1.2. 


ARITHMETIC. 

(PART  4.) 

PERCENTAGE. 

DEFINITIONS  AND  PRINCIPLES. 

1.  In  certain  operations,  particularly  those  pertaining  to 
business,  it  is  very  convenient  to  regard  the  quantity  on 
which  we  are  to  operate  as  being  divided  into  100  equal 
parts;  thus,  instead  of  using  the  ordinary  fractions  £,  f,  -f, 

28-^- 

we  use  the  equivalent  fractions  y2^,  T6¥°T,  — -^,  or  their  equiv- 
alent decimals,  .25,  .60,  .28^.  This  practice  is  a  very  con- 
venient one  in  all  computations  involving  United  States 
money,  because,  since  $1  equals  100  cents,  it  is  easier  to 
comprehend  what  part  of  the  whole  y%55  is  than  some  other 
equivalent  fraction,  asT*¥9^;  it  is  also  much  easier  to  com- 
pute with  fractions  whose  denominators  are  100  than  it  is  to 
compute  with  fractions  whose  denominators  are  composed  of 
other  figures. 

2.  Percentage  is  a  term  applied  to  those  arithmetical 
operations  in  which  the  number  or  quantity  to  be  operated 
upon  is  supposed  to  be  divided  into  100  equal  parts. 

3.  The  term  per  cent,  means  by  the  hundred.     Thus, 
8   per  cent,   of  a  number  means  8   hundredths;   i.  e.,  yf^, 
or  .08,   of  that  number;    8  per  cent,   of    250  is  250  X  y^, 
or  250  X  .08  —  20;  47  per  cent,  of  75  tons  is  75  X  tVV  =  75 
X  .47  =  35.25  tons.     The  statement  that  the  population  of 
a  city  has  increased  22  per  cent,  in  a  given  time,  say  from 

§  2 

For  notice  of  copyright,  see  page  immediately  folio  wing  the  title  page. 
H.  S.     I.— 5 


ARITHMETIC. 


1880  to  1890,  is  equivalent  to  saying  that  the  increase  is 
22  in  every  hundred ;  that  is,  for  every  100  in  1880  there 
are  22  more,  or  122,  in  1890. 

4:.  The  sign  of  per  cent,  is  $,  and  is  read  per  cent. 
Thus,  6$  is  read  six  per  cent.;  12|-$  is  read  twelve  and  one- 
half  per  cent.,  etc. 

5.  When  expressing  the  per  cent,  of  a  number  to  use  in 
calculations,  it  is  customary  to  express  it  decimally  instead 
of  fractionally.     Thus,  instead  of  expressing  6$,  25$,  and  43$ 
as  TW>  -oir>  and  T4A»  it  is  usual  to  express  them  as  .06,  .25, 
and  .43. 

6.  The  following  table  will  show  how  any  per  cent,  can 
be  expressed  either  as  a  decimal  or  as  a  fraction : 


1$ 

.01 

tfcr 

w 

.0025 

ior^ 

2$ 

.02 

•dhr  °r  A 

w 

.005 

|                    l 

5$ 

.05 

dhr  °r  A 

iw 

.015 

^°r* 

10$ 

.10 

TTTO  or  A 

6W 

.06i 

^or^ 

25$ 

.25 

T¥O  or  i 

8^ 

.08^ 

^  °r  A 

50$ 

.50 

TSA  or  i 

12** 

.125 

i^or^ 

75$ 

.75 

yVo  or  f 

18** 

•  16f 

lS°r- 

100$ 

1.00 

IHorl 

33W 

.33| 

li°r* 

125$ 

1.25 

W  or  li 

37i!< 

.37i 

lS°^ 

150$ 

1.50 

m  or  H 

62^ 

.625 

Ib10r* 

500$ 

5.00 

|^  or  5 

^ 

.875 

lS°r^ 

§  2  ARITHMETIC.  3 

7.  The  names  of  the  different  terms  used  in  percentage 
are :  the  base,  the  rate  or  rate  per  cent. ,  the  percentage,  the 
amount,  and  the  difference. 

8.  The  base  is  the  number  or  quantity  which  is  supposed 
to  be  divided  into  100  equal  parts. 

9.  The  rate  per  cent,  is  that  number  of  the  100  equal 
parts  into  which  the  base  is  supposed  to  be  divided  which 
is  taken  or  considered.     The  rate  is  the  number  of  hun- 
dredths   of    the   base   that   is   taken   or   considered.      The 
distinction  between  the  rate  per  cent,  and  the  rate  is  this: 
The  rate  per  cent,   is  always   100  times  the  rate.     Thus, 
7$  of  125  and  .07  of  125  amount  in  the  end  to  the  same 
thing;   the  former,  7,  is  the  rate  per  cent. — the  number  of 
hundredths  of  125  intended;  the  latter,  .07,  is  the  rate,  the 
part  of  125  that  is  to  be  found;  7$  is  used  in  speech,   .07 
is  the  form   used  in  computation.      So,    also,    12£$  =  .  125, 
\%  —  .005,  lf$  =  .0175.     In  the  table  just  given,  the  num- 
bers in  the  first  column  are  rates  per  cent. ;  those  in  the 
second  column  are  rates. 

10.  The  percentage  is  the  result  obtained  by  multiply- 
ing  the   base   by  the   rate.     Thus,   7$  of   125  —  125  X  .07 
=  8.75,  the  percentage. 

11.  The  amount  is  the  sum  of  the  base  and  the  per- 
centage. 

12.  The  difference  is  the  remainder  obtained  when  the 
percentage  is  subtracted  from  the  base. 

13.  The  terms   amount   and   difference   are   ordinarily 
used  when  there  is  an  increase  or  a  decrease  in  the  base. 
For  example,  suppose  the  population  of  a  village  is  1,500 
and  it  increases  25  per  cent.     This  means  that  for  every 
100  of  the  original  1,500  there  is  an  increase  of  25,  or  a  total 
increase   of   15  X  25  =  375.     This    increase    added   to   the 
original   population   gives   the  amount,    or  the    population 
after  the  increase.     If  the  population  had   decreased  375, 


4  ARITHMETIC.  §  2 

the  final  population  would  have  been  1,500  —  375=  1,125, 
and  this  would  be  the  difference.  The  original  popula- 
tion, 1,500,  is  the  base  on  which  the  percentage  is  com- 
puted; the  25  is  the  rate  per  cent.,  and  the  increase  or 
decrease,  375,  is  the  percentage.  If  the  base  increases,  the 
final  value  is  the  amount ;  and  if  it  decreases,  its  final  value 
is  the  difference. 


CALCULATIONS  INVOLVING  PERCENTAGE. 

14.  From  the  foregoing  it  is  evident  that  to  find  the 
percentage,  the  base  must  be  multiplied  by  the  rate.     Hence, 
the  following 

Rule. —  To  find  the  percentage,  multiply  the  base  by  the 
rate. 

EXAMPLE.— Out  of  a  lot  of  300  boiler  tubes  76$  was  used  in  a  boiler. 
How  many  tubes  were  used  ? 

SOLUTION.— The  rate  is  .76;  the  base  is  300;  hence,  the  number  of 
tubes  used,  or  the  percentage,  is  by  the  above  rule 

300  x- 76  =  228  tubes.     Ans. 
Expressing  the  rule  as  a 

Formula,  percentage  =  base  X  rate. 

15.  When  the  percentage  and  rate  are  given,  the  base 
may  be  found  by  dividing  the  percentage  by  the  rate.     For, 
suppose  that  12  is  6$,  or  yf^,  of  some  number;  then  1$,  or 
-j-J-g-,  of  the  number,  is  12  -r-  6,  or  2.     Consequently,  if  2  =  1$, 
or  Tfg-,   100$,   or   {$$  =  2  X  100  =  200.      But   as   the    same 
result  may  be  arrived  at  by  dividing  12  by  .06,  since  12  -H  .06 
=  200,  it  follows  that: 

Rule. —  When  the  percentage  and  rate  are  given,  to  find  the 
base,  divide  the  percentage  by  the  rate. 

Formula,  base  =  percentage  -=-  rate. 


§  2  ARITHMETIC.  5 

EXAMPLE. —  76$  of  a  lot  of  boiler  tubes  was  used  in  the  construction 
of  a  boiler.  If  the  number  of  tubes  used  was  228,  how  many  tubes  were 
in  the  lot  ? 

SOLUTION. — Here  228  is  the  percentage  and  .76  is  the  rate;  hence, 
applying  the  rule, 

228  -j-  .  76  =  300  tubes.     Ans. 

16.  When  the  base  and  percentage  are  given,  to  find  the 
rate,  the  rate  may  be  found  by  dividing  the  percentage  by 
the  base.  For,  suppose  that  it  is  desired  to  find  what  per 
cent.  12  is  of  200,  \%  of  200  is  200  X  .01  =  2.  Now,  if  1# 
is  2,  12  is  evidently  as  many  per  cent,  as  2  is  contained 
times  in  12,  or  12  -4-  2  =  6#.  But  the  same  result  may  be 
obtained  by  dividing  12,  the  percentage,  by  200,  the  base, 
since  12  -j-  200  =  .06  =  6#.  Hence, 

Rule. —  When  the  percentage  and  base  are  given,  to  find  the 
rate,  divide  the  percentage  by  the  base,  and  the  result  will  be 
the  rate. 

Formula,  rate  =  percentage  -r-  base. 

EXAMPLE  1.— Out  of  a  lot  of  300  boiler  tubes  228  were  used.  What 
per  cent,  of  the  total  number  was  used  ? 

SOLUTION. — Here  300  is  the  base  and  228  is  the  percentage ;  hence, 
applying  rule, 

Rate  =  228  -f-  300  =  .  76  =  76*.     Ans. 

EXAMPLE  2.— What  per  cent,  of  875  is  25  ? 

SOLUTION.— Here  875  is  the  base  and  25  is  the  percentage;  hence, 
applying  rule, 

Rate  =  25  H-  875  =  .02$  =  2f*.     Ans. 

PROOF.—    875  X  .024  =  25. 


EXAMPLES   FOB  PRACTICE. 

17.      What  per  cent  of: 


(a)  360  is  90  ? 

\b)  900  is  360  ? 

(c)  125  is  25  ? 

(</)  150  is  750  ? 

(e)  280  is  112  ? 

(/)  400  is  200? 

(g)  47  is  94? 

(fa  500  is  250  ? 


Ans. 


(a)  25*. 

(6)  40*. 

(c)  20*. 

(d)  500*. 

(e)  40*. 
(/)  50*. 
(g)  200*. 
(k)  50*. 


6  ARITHMETIC.  §  2 

18.  The  amount  may  be  found,  when  the  base  and  rate 
are   given,    by   multiplying   the   base   by  1    plus   the  rate 
expressed  decimally.     For,  suppose  that  it  is  desired  to  find 
the  amount  when  200  is  the  base  and  .06  is  the  rate.     The 
percentage   is  200  X  .06  =12,  and,  according  to  definition, 
Art.   11,  the  amount  is  200  +  12  =  212.      But  the   same 
result  may  be  obtained  by  multiplying  200  by  1  -f  .06,  or  1.06, 
since  200  X  1.06  =  212.    Hence, 

Rule. —  When  the  base  and  rate  are  given,  to  find  the 
amount,  multiply  the  base  by  1  plus  the  rate. 

Formula,  amount  =  base  X  (1  +  rate). 

EXAMPLE. — If  a  man  earned  $725  in  a  year,  and  the  next  year  10$ 
more,  how  much  did  he  earn  the  second  year  ? 

SOLUTION. — Here  725  is  the  base  and  .10  is  the  rate,  and  the  amount 
is  required.  Hence,  applying  the  rule, 

725X1. 10  =  $797. 50.     Ans. 

19.  When  the  base  and  rate  are  given,  the  difference 
may  be  found  by  multiplying  the  base  by  1  minus  the  rate 
expressed  decimally.     For,  suppose  that  it  is  desired  to  find 
the  difference  when  the  base  is  200  and  the  rate  is  6$.     The 
percentage  is  200  X  .06  =  12;  and,  according  to  definition, 
Art.   12,  the  difference  =  200  —  12  =  188.      But  the  same 
result  may  be  obtained  by  multiplying  200  by  1  —  .06,  or.  94, 
since  200  X  .94  =  188.     Hence, 

Rule. —  When  the  base  and  rate  are  given,  to  find  the 
difference,  multiply  the  base  by  1  -minus  the  rate. 

Formula,  difference  =  base  X  (1  —  rate). 

EXAMPLE.— Out  of  a  lot  of  300  boiler  tubes  all  but  24£  were  used  in 
one  boiler ;  how  many  tubes  were  used  ? 

SOLUTION.— Here  300  is  the  base,  .24  is  the  rate,  and  it  is  desired  to 
find  the  difference.  Hence,  applying  the  rule, 

300  X  (1  -  -24)  =  228  tubes.     Ans. 

20.  When  the  amount  and  rate  are  given,  the  base  may 
be  found  by  dividing  the  amount  by  1  plus  the  rate.     For, 
suppose  that  it  is   known   that   212   equals   some   number 


§  2  ARITHMETIC.  7 

increased  by  6$  of  itself.  Then,  it  is  evident  that  212 
equals  106$  of  the  number  (base)  that  it  is  desired  to  find. 

f\  -|  tf) 

Consequently,    if   212  =  106$,    1$  =  ^~  =  2,  and   100$  =  2 

X  100  =  200  =  the  base.  But  the  same  result  may  be 
obtained  by  dividing  212  by  1  +  .06  or  1.06,  since  212  -h  1.06 
=  200.  Hence, 

Rule. —  When  the  amount  and  rate  are  given,  to  find  the 
base,  divide  the  amount  by  1  plus  the  rate. 

Formula,  base  =  amount  -4-  (1  -|-  rate). 

EXAMPLE. — The  theoretical  discharge  of  a  certain  pump  when  run- 
ning at  a  piston  speed  of  100  feet  per  minute  is  278,910  gallons  per  day 
of  10  hours.  Owing  to  leakage  and  other  defects,  this  value  is  25£ 
greater  than  the  actual  discharge.  What  is  the  actual  discharge  ? 

SOLUTION. — Here  278,910  equals  the  actual  discharge  (base)  increased 
by  25#  of  itself.  Consequently,  278,910  is  the  amount  and  25£  is  the 
rate.  Applying  rule, 

Actual  discharge  =  278,910  H-  1.25  =  223,128  gal.     Ans. 

21.  When  the  difference  and  rate  are  given,  the  base 
may  be  found  by  dividing  the  difference  by  1  minus  the 
rate.  For,  suppose  that  188  equals  some  number  less  6$  of 
itself.  Then,  188  evidently  equals  100  —  6  =  94$  of  some 
number.  Consequently,  if  188  =  94$,  1$  =  188  -4-  94  =  2, 
and  100$  =  2  X  100  =  200.  But  the  same  result  may  be 
obtained  by  dividing  188  by  1  —  .06,  or  .94,  since  188  -4-  .94 
=  200.  Hence, 

Rule. —  When  the  difference  and  rate  are  given,  to  find  the 
base,  divide  the  difference  by  1  minus  the  rate. 

Formula,  base  =  difference  ~  (1  —  rate]. 

EXAMPLE  1.— From  a  lot  of  boiler  tubes  76#  was  used  in  the  constrtic- 
tion  of  a  boiler.  If  there  were  72  tubes  unused,  how  many  tubes  were 
in  the  lot  ? 

SOLUTION. — Here  72  is  the  difference  and  .76  is  the  rate.  Applying 
rule, 

73  ^_  (i  _  .76)  =  300  tubes.     Ans. 


8  ARITHMETIC.  §  2 

EXAMPLE  2. — The  theoretical  number  of  foot-pounds  of  work  per  min- 
ute required  to  operate  a  boiler  feed-pump  is  127,344.  If  30$  of  the 
total  number  actually  required  be  allowed  for  friction,  leakage,  etc., 
how  many  foot-pounds  are  actually  required  to  work  the  pump  ? 

SOLUTION. — Here  the  number  actually  required  is  the  base;  hence, 
127,344  is  the  difference  and  .30  is  the  rate.  Applying  the  rule, 

127,344  -4-  (1  —  .30)  =  181,920  foot-pounds.     Ans. 

22.  EXAMPLE. — A  certain  chimney  gives  a  draft  of  2.76  inches  of 
water.     By  increasing  the  height  20  feet,  the  draft  was  increased  to 
3  inches  of  water.     What  was  the  gain  per  cent.? 

SOLUTION. — Here  it  is  evident  that  3  inches  is  the  amount  and  that 
2.76  inches  is  the  base.  Consequently,  3  —  2.76  =  .24  inch  is  the  per- 
centage, and  it  is  required  to  find  the  rate.  Hence,  applying  the  rule 
given  in  Art.  16, 

Gain  per  cent.  =  .24  -r-  2.76  =  .087  =  8.7#.     Ans. 

23.  EXAMPLE.— A  certain  chimney  gave  a  draft  of  3  inches  of 
water.     After  an  economizer  had  been  put  in,  the  draft  was  reduced  to 
1.2  inches  of  water.     What  was  the  loss  per  cent.  ? 

SOLUTION. — Here  it  is  evident  that  1.2  inches  is  the  difference  (since 
it  equals  3  inches  diminished  by  a  certain  per  cent,  loss  of  itself)  and 
3  inches  is  the  base.  Consequently,  3  —  1.2  =  1.8  inches  is  the  percent- 
age. Hence,  applying  the  rule  given  in  Art.  16, 

Loss  per  cent.  =1.8  -j-  3  =  .60  =  60#.     Ans. 

24.  To  flnd  the  gain  or  loss  per  cent.  : 

Rule. — Find  the  difference  between  the  initial  and  the  final 
value ;  divide  this  difference  by  the  initial  value. 

EXAMPLE.— If  a  man  buys  a  steam  engine  for  $1,860  and  some  time 
afterwards  purchases  a  condenser  for  25#  of  the  cost  of  the  engine,  does 
he  gain  or  lose,  and  how  much  per  cent.,  if  he  sells  both  engine  and 
condenser  for  $2,100  ? 

SOLUTION.— The  cost  of  the  condenser  was  $1,860  X  .25  =  $465;  con- 
sequently, the  initial  value,  or  cost,  was  $1,860  +  $465  =  $2,325.  Since 
he  sold  them  for  $2,100,  he  lost  $2,325  -  $2,100  =  $225.  Hence,  apply- 
ing rule, 

225  H-  2, 325  =  .  0968  =  9. 68£  loss.     Ans. 


2  ARITHMETIC. 

EXAMPLES  FOR  PRACTICE. 

25.     Solve  the  following: 

(a)  What  is  12|$  of  $900  ? 

(6)  What  is  #  of  627  ? 

(0  What  is  33£$  of  54  ? 

(^/)  101  is  68f$  of  what  number  ? 


(e)  784  is  83£$  of  what  number  ? 

(/)  What  $  of  960  is  160  ? 

( £•)  What  $  of  $3, 606  is  $450f  ? 

(//)  What  $  of  280  is  112  ? 


(a)  $112.50. 

(b)  5.016. 
(<:)      18. 


940.8. 


40$. 


1.  A  steam  plant  consumed  an  average  of  3,640  pounds  of  coal  per 
day.    The  engineer  made  certain  alterations  which  resulted  in  a  saving 
of  250  pounds  per  day.     What  was  the  per  cent,  of  coal  saved  ? 

Ans.  7$,  nearly. 

2.  If  the  speed  of  an  engine  running  at  126  revolutions  per  minute 
should  be  increased  64-$,  how  many  revolutions  per  minute  would  it 
then  make?  Ans.  134.19  rev. 

3.  The  list  price  of  an  engine  was  $1,400 ;  of  a  boiler,  $1,150;  and  of 
the  necessary  fittings  for  the  two,  $340.     If  25$  discount  was  allowed 
on  the  engine,  22$  on  the  boiler,  and  12^$  on  the  fittings,  what  was  the 
actual  cost  of  the  plant?  Ans.  $2,244.50. 

4.  If  I  lend  a  man  $1,100,  and  this  is  18#  of  the  amount  that  I  have 
on  interest,  how  much  money  have  I  on  interest  ?  Ans.  $5,945.95. 

5.  A  test  showed  that  an  engine  developed  190.4  horsepower,  15$  of 
which  was  consumed  in  friction.     How  much  power  was  available  for 
use?  Ans.  161.84  H.  P. 

6.  By  adding  a  condenser  to  a  steam  engine,  the  power  was  increased 
14$  and  the  consumption  of  coal  per  horsepower  per  hour  was  decreased 
20$.    If  the  engine  could  originally  develop  50  horsepower  and  required 
3£  pounds  of  coal  per  horsepower  per  hour,  what  would  be  the  total 
weight  of  coal  used  in  an  hour  with  the  condenser,  assuming  the  engine 
to  run  full  power  ?  Ans.  159.6  Ib. 


DE^OMI^ATE    LUMBERS. 

26.  A  denominate  number  is  a  concrete  number,  and 
may  be  either  simple  or  compound;  as,  8  quarts;  5  feet 
10  inches,  etc.     Denominate  numbers  are  also  called  com- 
pound numbers. 

27.  A  simple  denominate  number  consists  of  units  of 
but  one  denomination;  as,  16  cents;  10  hours;  5  dollars,  etc. 


10  ARITHMETIC.  §  2 

28.  A  compound  denominate  number  consists  of  units 
of  two  or  more  denominations  of  a  similar  kind ;  as,  3  yards 
2  feet  1  inch. 

29.  In  whole  numbers  and  in  decimals,  the  law  of  increase 
and  decrease  is  on  the  scale  of  10,  but  in  compound,  or  denom- 
inate, numbers  the  scale  varies. 


MEASURES. 

30.  A  measure  is  a  standard  unit,  established  by  law  or 
custom,  by  which   quantity  of  any  kind   is  measured.     The 
standard  unit  of  dry  measure  is  the  Winchester  bushel ;  of 
•weight,  the  pound ;  of  liquid  measure,  the  gallon,  etc. 

31.  Measures  are  of  six  kinds: 

1.  Extension.  4.     Time. 

2.  Weight.  5.     Angles. 

3.  Capacity.  6.     Money  or  value. 


MEASURES    OF    EXTENSION. 

32.     Measures    of    extension    are  used    in    measuring 
lengths,  distances,  surfaces,  and  solids. 


LIXEAR  MEASURE. 

TABLE. 
Abbreviation. 


12  inches  (in.)  =  1  foot  .  .  ft. 

3  feet  .  .  .  =  1  yard .  .  yd. 

5.5  yards    .     .  =  1  rod    .     .  rd. 

40  rods    .     .     .  =  1  furlong .  fur. 

8  furlongs  .     .  =  1  mile  .     .  mi. 


in.  ft.  yd.          rd.  fur.  mi. 

36=   3      =1 
198  =  16|     =5.5      =      1 
7,920  =  660     =220     =    40=1 
63,360  =  5,280  =  1,760  =  320=8  =  1 


SQUARE  MEASURE. 

TABLE. 

144    square  inches  (sq.  in.)    .     .     .  =  1  square  foot      .     .     .     .     sq.  ft. 

9    square  feet =  1  square  yard     .     .     .     .   sq.  yd. 

30£  square  yards =  1  square  rod sq.  rd. 

160    square  rods =1  acre A. 

640    acres =  1  square  mile      .     .     .     .   sq.  mi. 

sq.  mi.        A.  sq.  rd.          sq.  yd.  sq.  ft.  sq.  in. 

1     =    640  =    102,400  =  3,097,600    =    27,878,400  =  4,014,489,600 


§  2  ARITHMETIC.  11 

CUBIC  MEASURE. 

TABLE. 

1,728    cubic  inches  (cu.  in.)  .     .  =     1  cubic  foot cu.  ft 

27    cubic  feet =1  cubic  yard cu.  yd. 

128    cubic  feet =     1  cord cd. 

24f  cubic  feet =     1  perch P. 

cu.  yd.    cu.  ft.        cu.  in. 
1     =     27     =     46,656 


MEASURES  OF  WEIGHT. 
AVOIRDUPOIS   WEIGHT. 

TABLE. 

16  ounces  (oz.) =     1  pound Ib. 

100  pounds =     1  hundredweight      .     .     .      cwt. 

20  cwt.,  or  2,000  Ib =     1  ton T. 

T.          cwt.  Ib.  oz. 

1     =     20     =     2,000     =     32,000 

33.  The   ounce   is   divided   into  halves,    quarters,    etc. 
Avoirdupois  weight  is  used  for  weighing  coarse  and  heavy 
articles. 

LONG   TON  TABLE. 

16  ounces =     1  pound Ib. 

112  pounds =     1  hundredweight      .     .     .     cwt. 

20  cwt.,  or  2,240  Ib =     1  ton T. 

34.  In  all  the  calculations  hereafter,  2,000  pounds  will 
be  considered  1  ton,  unless  the  long  ton  (2,240  pounds)  is 
especially  mentioned. 

TROY    WEIGHT. 

TABLE. 

24  grains  (gr.) =     1  pennyweight     ....     pwt. 

20  pennyweights =     1  ounce qz. 

12  ounces =1  pound Ib. 

Ib.          oz.  pwt.  gr. 

1     =     12     =     240     =     5,760 

Troy  weight  is  used  in  weighing  gold  and  silver  ware, 
jewels,  etc.      It  is  used  by  jewelers. 


12  ARITHMETIC.  §  2 

MEASURES   OF  CAPACITY. 
LIQUID  MEASURE. 

TABLE. 

4    gills  (gi.) =  1  pint pt. 

2    pints =  1  quart qt. 

4    quarts =  1  gallon gal. 

31|  gallons =  1  barrel bbl. 

2    barrels,  or  63  gallons    .     .     .  =  1  hogshead hhd. 

hhd.      bbl.        gal.          qt.  pt.  gi. 

1     =     2     =     63     =     252     =     504     =     2,016 

DRY  MEASURE. 

TABLE. 

2  pints  (pt.) =     1  quart qt. 

8  quarts      ........     =     1  peck pk. 

4  pecks =     1  bushel bu. 

bu.         pk.         qt.  pt. 

1     =     4     =     32     =     64 


MEASURE  OF  TIME. 

TABLE. 

60  seconds  (sec.) =     1  minute min. 

60  minutes =     1  hour hr. 

24  hours =     1  day da. 

7  days =1  week wk. 

365  days       \    .  .     =     1  common  year  .  .  yr. 
12  months  ) 

366  days =     1  leap  year. 

100  years =    1  century. 

NOTE. — It  is  customary  to  consider  one  month  as  30  days. 


MEASURE  OF  ANGLES  OR  ARCS. 

TABLE. 

60  seconds  (") =     1  minute '. 

60  minutes =     1  degree °. 

90  degrees =     1  right  angle  or  quadrant  |_. 

360  degrees =1  circle cir. 

1  cir.     =    360°     =     21,600'     =     1,296,000' 


ARITHMETIC.  13 

MEASURE    OF    MONEY. 
UNITED  STATES  MOXEY. 


10  cents    

—     1  dime  

10  dimes    

_     j 

dollar     . 

10  dollars 

.. 

E. 

$ 

d. 

ct. 

m. 

1     = 

10 

=      100     = 

1,000    = 

10,000 

MISCELLANEOUS    TABLE. 


12  things  are  1  dozen. 

12  dozen  are  1  gross. 

12  gross  are  1  great  gross. 

2  things  are  1  pair. 
20  things  are  1  score. 

1  league  is  3  miles. 

1  fathom  is  6  feet. 


1  meter  is  39.37  inches. 

1  hand  is  4  inches. 

1  palm  is  3  inches. 

1  span  is  9  inches. 
24  sheets  are  1  quire. 
20  quires,  or  480  sheets,  are  1  ream. 

1  bushel  contains  2, 150.4  cu.  in. 


1  U.  S.  standard  gallon  (also  called  a  wine  gallon)  contains  231  cu.  in. 
1  U.  S.  standard  gallon  of  water  weighs  8.355  pounds,  nearly. 
1  cubic  foot  contains  7.481  U.  S.  standard  gallons,  nearly. 
1  British  imperial  gallon  of  water  weighs  10  pounds. 

It  will  be  of  great  advantage  to  the  student  to  carefully 
memorize  all  the  above  tables. 


REDUCTION  OF  DENOMINATE  NUMBERS. 

35.  Reduction  of  denominate  numbers  is  the  process  of 
changing  their  denomination  without  changing  their  value. 
They  may  be  changed  from  a  higher  to  a  lower  denomina- 
tion or  from  a  lower  to  a  higher — either  is  reduction.     As 

2  hours  =  120  minutes. 
32  ounces  =  2  pounds. 

36.  Principle. — Denominate    numbers  are    changed  to 
lower  denominations  by  multiplying,  and  to  higher  denomi- 
nations by  dividing. 


14  ARITHMETIC.  §  2 

To  reduce  denominate  numbers  to  lower  denomina- 
tions : 

37.     EXAMPLE.— Reduce  5  yd.  2  ft.  7  in.  to  inches. 
SOLUTION. —  yd.  ft.  in. 


15ft. 
2ft. 

FT"  ft. 

12 

3~4 
17 


2  0  4  in. 

7  in. 

'  iTT  in.     Ans. 

EXPLANATION. — Since  there  are  3  feet  in  1  yard,  in  5  yards 
there  are  5  X  3,  or  15,  feet,  and  15  feet  +  2  feet  —  17  feet. 
There  are  12  inches  in  a  foot ;  therefore,  12  X  17  =  204  inches, 
and  204  inches  -f-  7  inches  =  211  inches  in  5  yards  2  feet 
7  inches. 

38.     EXAMPLE. — Reduce  6  hours  to  seconds. 
SOLUTION. —  6        hr. 

60 

360     min. 


21600  sec.     Ans. 

EXPLANATION. — As  there  are  60  minutes  in  1  hour,  in 
6  hours  there  are  6  X  60,  or  360,  minutes;  as  there  are  no 
minutes  to  add,  we  multiply  360  minutes  by  60,  to  get  the 
number  of  seconds. 

39.  In  order  to  avoid  mistakes,  if  any  denomination  be 
omitted,    represent  it  by  a  cipher.     Thus,  before  reducing 
3  rods  6  inches  to  inches,  insert  a  cipher  for  yards  and  a 
cipher  for  feet,  as 

rd.     yd.      ft.       in. 
3006 

40.  Rule. — Multiply  the  number  representing  the  highest 
denomination  by  the  number  of  units  in  the  next  lower  required 


§  2  ARITHMETIC.  15 

to  make  one  of  tJie  liiglier  denomination,  and  to  t/ie  product 
add  the  number  of  given  units  of  that  lower  denomination. 
Proceed  in  this  manner  until  the  number  is  reduced  to  the 
required  denomination. 


EXAMPLES   FOR  PRACTICE. 
41.     Reduce: 
(a)     4  rd.  2  yd.  2  ft.  to  feet.  f  (a)     74  ft. 


(b)  4  bu.  3  pk.  2  qt.  to  quarts. 

(c)  13  rd.  5  yd.  2  ft.  to  feet. 

(d)  5  mi.  100  rd.  10  ft.  to  feet.  Ans. 

(e)  52  hhd.  24  gal.  1  pt.  to  pints. 
(/)  5  cir.  16°  20'  to  minutes. 

(g)  14  bu.  to  quarts. 


(6)  154  qt. 

(c)  231.5  ft. 

(d)  28,060ft. 

(e)  26,401  pt. 


(g)    448  qt. 


To  reduce  lower  to  higher  denominations : 

42.  EXAMPLE. — Reduce  211  inches  to  higher  denominations: 
SOLUTION.—  12)211  in. 

3  )  1  7  ft.  +  7  in. 

5  yd.  +  2  ft.  +  7  in.     Ans. 

EXPLANATION. — There  are  12  inches  in  1  foot;  therefore, 
211  divided  by  12  =  17  feet  and  7  inches  over.  There  are 
3  feet  in  1  yard;  therefore,  17  feet  divided  by  3  =  5  yards 
and  2  feet  over.  The  last  quotient  and  the  two  remainders 
constitute  the  answer,  5  yards  2  feet  7  inches. 

43.  EXAMPLE. — Reduce  14,135  gills  to  higher  denominations. 
SOLUTION.—  4)  14135 

2  )   3533  pt.  3  gi. 

4)   1766qt.  1  pt. 

441  gal.  2  qt. 

31.5)441.0(14  bbl. 
315 


1260 
1260 

EXPLANATION. — There  are  4  gills  in  1  pint,  and  in  14,135 
gills  there  are  as  many  pints  as  4  is  contained  times  in 
14,135,  or  3,533  pints  and  3  gills  remaining.  There  are 
2  pints  in  1  quart,  and  in  3,533  pints  there  are  1,766  quarts 
and  1  pint  remaining.  There  are  4  quarts  in  1  gallon,  and 


16  ARITHMETIC.  §  2 

in  1,766  quarts  there  are  441  gallons  and  2  quarts  remain- 
ing. There  are  31^  gallons  in  1  barrel,  and  in  441  gallons 
there  are  14  barrels. 

The  last  quotient  and  the  three  remainders  constitute 
the  answer,  14  barrels  2  quarts  1  pint  3  gills. 

44.  Rule. — Divide  the  number  representing  the  denom- 
ination given  by  the  number  of  units  of  this  denomination 
required  to  make  one  unit  of  the  next  higher  denomination. 
The  remainder  will  be  of  the  same  denomination,  but  the 
quotient  will  be  of  the  next  higher.  Divide  this  quotient  by 
the  number  of  units  of  its  denomination  required  to  make 
one  unit  of  the  next  higher.  Continue  iintil  the  highest 
denomination  is  reached  or  until  there  is  not  enough  of  a 
denomination  left  to  make  one  of  the  next  higher.  The  last 
quotient  and  the  remainders  constitute  the  required  result. 


EXAMPLES  FOR  PRACTICE. 

45.      Reduce  to  units  of  higher  denominations: 
(a)  7,460  sq.  in.;  (£)  7,580  sq.  yd.;  (c)  148,760  cu.  in.;  (d)  17,651"; 
(e)  8,000  gi. ;  (/)  36,450  Ib. 

(a)      5  sq.  yd.  6  sq.  ft.  116  sq.  in. 

(6)      1  A.  90  sq.  rd.  17  sq.  yd.  4  sq.  ft.  72  sq.  in. 

(<;)      3  cu.  yd.  5  cu.  ft.  152  cu.  in. 

(d)  4°  54'  11". 

(e)  3  hhd.  61  gal. 
(/)    18  T.  4  cwt.  50  Ib. 


Ans. 


ADDITION  Or  DENOMINATE  NUMBERS. 

46.  As  in  the   case  of    abstract   numbers,   denominate 
numbers  may  be  added,  subtracted,  multiplied,  and  divided. 

47.  EXAMPLE.— Find  the  sum  of  3  cwt.  46  Ib.  12  oz. ;  8  cwt.  12  Ib. 
13  oz. ;  12  cwt.  50  Ib.  13  oz. ;  27  Ib.  4  oz. 

SOLUTION. —             T.  cwt.  Ib.  oz. 

0             3  46  12 

0             8  12  13 

0  12  50  13 

0              0  27  4 


37  10    Ans. 


§  2  ARITHMETIC.  17 

EXPLANATION. — Begin  to  add  at  the  right-hand  column; 
4  -f- 13  +  13  +  12  =  42  ounces;  as  16  ounces  make  1  pound, 
42  ounces  -=-  16  =  2  pounds  and  a  remainder  of  10  ounces,  or 
2  pounds  and  10  ounces.  Place  10  ounces  under  the  ounce 
column  and  add  2  pounds  to  the  next,  or  pound,  column. 
Then,  2  +  27  +  50  +  12  -f  46  =  137  pounds;  as  100  pounds 
make  a  hundredweight,  137  -4-  100  =  1  hundredweight  and 
a  remainder  of  37  pounds.  Place  the  37  under  the  pound 
column  and  add  1  hundredweight  to  the  next,  or  hundred- 
weight, column.  Next,  1  -)-  12  +  8  +  3  =  24  hundredweight. 
20  hundredweight  make  a  ton ;  therefore,  24  -^  20  =  1  ton 
and  4  hundredweight  remaining.  Hence,  the  sum  is  1  ton 
4  hundredweight  37  pounds  10  ounces. 

48.      EXAMPLE. — What  is  the  sum  of  2  rd.  3  yd.  2  ft.  5  in. ;  6  rd. 

1  ft.  10  in. ;  17  rd.  11  in. ;  4  yd.  1  ft.? 
SOLUTION. — 


rd. 

yd. 

ft. 

in. 

2 

3 

2 

5 

6 

0 

1 

10 

17 

0 

0 

11 

0 

4 

1 

0 

26  8i          0  2 

or    26  3  1  8    Ans. 

EXPLANATION. — The  sum  of  the  numbers  in  the  first  col- 
umn =  26  inches,  or  2  feet  and  2  inches  remaining.  The 
sum  of  the  numbers  in  the  next  column  plus  2  feet  =:  6  feet, 
or  2  yards  and  0  feet  remaining.  The  sum  of  the  numbers 
in  the  next  column  plus  2  yards  =  9  yards,  or  9  -f-  5£  =  1  rod 
and  3£  yards  remaining.  The  sum  of  the  next  column  plus 
1  rod  =  26  rods.  To  avoid  fractions  in  the  sum,  the  %  yard 
is  reduced  to  1  foot  and  6  inches,  which  added  to  26  rods 
3  yards  0  feet  and  2  inches  =  26  rods  3  yards  1  foot  8  inches. 

49.  EXAMPLE.— What  is  the  sum  of  47  ft.  and  3  rd.  2  yd.  2  ft. 
10  in.? 

SOLUTION. — When  47  ft.  is  reduced,  it  equals  2  rd.  4  yd.  2  ft.,  which 
can  be  added  to  3  rd.  2  yd.  2  ft.  10  in.     Thus, 
rd.         yd.         ft.         in. 
3  2  2          10 

2420 


6  H          1          10 

6204    Ans. 


H.  8.    L—6 


18  ARITHMETIC.  §  2 

5O.  Rule. — Place  the  numbers  so  that  like  denominations 
are  under  one  another.  Begin  at  the  right-hand  column  and 
add.  Divide  the  sum  by  the  number  of  units  of  this  denomi- 
nation required  to  make  one  unit  of  the  next  higher.  Place 
the  remainder  under  the  column  added  and  carry  the  quotient 
to  the  next  column.  Continue  in  this  manner  until  the  highest 
denomination  given  is  reached. 


EXAMPLES  FOR  PRACTICE. 

51.     What  is  the  sum  of: 

(a)  25  Ib.  7  oz.  15  pwt.  23  gr. ;  17  Ib.  16  pwt. ;  15  Ib.  4  oz.  12  pwt. ; 
18  Ib.  16  gr. ;  10  Ib.  2  oz.  11  pwt.  16  gr.  ? 

(b)  9  mi.  13  rd.  4  yd.  2  ft.  ;  16  rd.  5  yd.  1  ft.  5  in.  ;  16  mi.  2  rd.  3  in.  ; 
14  rd.  1  yd.  9  in.  ? 

(c)  3  cwt.  46  Ib.  12  oz. ;  12  cwt.  9|  Ib. ;  2|  cwt.  21$  Ib.  ? 

(d)  10  yr.  9  mo.  1  wk.  3  da. ;  42  yr.  6  mo.  7  da. ;  7  yr.  9  mo.  2  wk. 
4  da. ;  17  yr.  17  da.  ? 

(e)  17  T.  11  cwt.  49  Ib.  14  oz..;  16  T.  47  Ib.  13  oz. ;  20  T.  13  cwt.  14  Ib. 
6  oz. ;  11  T.  4  cwt.  16  Ib.  12  oz.  ? 

(/)  14  sq.  yd.  8  sq.  ft.  19  sq.  in. ;  105  sq.  yd.  16  sq.  ft.  240  sq.  in. ; 
42  sq.  yd.  28  sq.  ft.  165  sq.  in.  ? 

(a)  86  Ib.  3  oz.  16  pwt.  7  gr. 

(b)  25  mi.  47  rd.  1  ft.  5  in. 
,  (c)   18  cwt.  2  Ib.  14  oz. 

4  (d)  78  yr.  1  mo.  3  wk.  3  da. 
(e)  65  T.  9  cwt.  28  Ib.  13oz. 
(/)  167  sq.  yd.  136  sq.  in. 


SUBTRACTION   OF   DENOMINATE   NUMBERS. 

52.  EXAMPLE.— From  21  rd.  2  yd.  2  ft.  6*  in.  take  9  rd.  4  yd. 
10i  in. 

SOLUTION.—  rd.         yd.         ft.          in. 

21  2  2  6i 

9  4  0          10J 

11  34          1  8i    Ans. 

EXPLANATION. — Since  10^  inches  cannot  be  taken  from 
6£  inches,  we  must  borrow  1  foot,  or  12  inches,  from  the 
2  feet  in  the  next  column  and  add  it  to  the  6£.  6$  +  12 
=  18$.  18$  inches  —  10±  inches  =  8£  inches.  Then,  0  foot 


§  2  ARITHMETIC.  19 

from  the  1  remaining  foot  =  1  foot.  4  yards  cannot  be  taken 
from  2  yards ;  therefore,  we  borrow  1  rod,  or  5£  yards,  from 
21  rods  and  add  it  to  2.  2  +  5£  =  7|;  7£  —  4  =  3£  yards. 

9  rods  from  20   rods  —  11  rods.     Hence,   the  remainder  is 
11  rods  3|-  yards  1  foot  8^  inches. 

To  avoid  fractions  as  much  as  possible,  we  reduce  the 
£  yard  to  inches,  obtaining  18  inches ;  this  added  to  8  J  inches 
gives  26£  inches,  which  equals  2  feet  2£  inches.  Then,  2  feet 
-f- 1  foot  =  3  feet  =  1  yard,  and  3  yards  -f-  1  yard  =  4  yards. 
Hence,  the  above  answer  becomes  11  rods  4  yards  0  feet 
2£  inches. 

53.  EXAMPLE. — What  is  the  difference  between  3  rd.  2  yd.  2  ft. 

10  in.  and  47  ft.? 

SOLUTION.—    47  ft.  =  2  rd.  4  yd.  2  ft. 

rd.  yd.          ft.          in. 

3  2  2  10 

2 4 2 0 

0  3£  0  10 

or  324    Ans. 

To  find  (approximately)  the  Interval  of  time  between 
two  dates : 

54.  EXAMPLE.  —  How    many    years,    months,    days,    and    hours 
between  4  o'clock  p.  M.  of  June  15,  1868,  and  10  o'clock  A.  M.,  Sep- 
tember 28,  1891 ? 

SOLUTION. —  yr.  mo.        da.  hr. 

1891  9  28  10 

1868  6  15  16 

23  3  12  18    Ans. 

EXPLANATION. — Counting  24  hours  in  1  day,  4  o'clock  p.  M. 
is  the  16th  hour  from  the  beginning  of  the  day,  or  midnight. 
September  is  the  9th  month  and  June  is  the  6th  month. 
After  placing  the  earlier  date  under  the  later  date,  subtract 
as  in  the  previous  problems.  Count  30  days  as  1  month. 

55.  Rule. — Place  the  smaller  quantity  under  the  larger 
quantity,  with  like  denominations  under  each  other.  Beginning 
at  tJie  right,  subtract  successively  the  number  in  the  subtrahend 
in  each  denomination  from  the  one  above  and  place  the  dif- 
ferences underneath.     If  the  number  in  the  minuend  of  any 


20  ARITHMETIC.  §  2 

denomination  is  less  than  the  number  under  it  in  the  sub- 
traliend,  one  unit  must  be  borrowed  from  the  minuend  of  the 
next  higher  denomination,  reduced,  and  added  to  it. 


EXAMPLES  FOR  PRACTICE. 
56.     From : 

(a)  125  Ib.  8  oz.  14  pwt.  18  gr.  take  96  Ib.  9  oz.  10  pwt.  4  gr. 
(£)  126  hhd.  27  gal.  take  104  hhd.  14  gal.  1  qt.  1  pt. 

(c)  65  T.  14  cwt.  64  Ib.  10  oz.  take  16  T.  11  cwt.  14  oz. 

(d)  148  sq.  yd.  16  sq.  ft.  142  sq.  in.  take  132  sq.  yd.  136  sq.  in. 
0?)  100  bu.  take  28  bu.  2  pk.  5  qt.  1  pt. 

(/)  14  mi.  36  rd.  5  yd.  13  ft.  11  in.  take  3  mi.  29  rd.  4  ft.  10  in. 

(a)  28  Ib.  11  oz.  4  pwt.  14  gr. 

(b)  22  hhd.  12  gal.  2  qt.  1  pt. 
(c}  49  T.  3  cwt.  63  Ib.  12  oz. 

(d)  16  sq.  yd.  16  sq.  ft.  6  sq.  in. 

(e)  71  bu.  1  pk.  2  qt.  1  pt. 
(/)  11  mi.  7  rd.  5  yd.  9  ft.  1  in. 


MULTIPLICATION  OF  DENOMINATE  NUMBERS. 

57.      EXAMPLE.— Multiply  7  Ib.  5  oz.  13  pwt.  15  gr.  by  12. 
SOLUTION.—  Ib.         oz.        pwt.         gr. 

7  5  13  15 

12 


89  8  3  12    Ans. 

EXPLANATION. —  15  grains  X  12  =  180  grains.  180  -^  24 
=  7  pennyweights  and  12  grains  remaining.  Place  the  12  in 
the  grain  column  and  carry  the  7  pennyweights  to  the  next 
column.  Now,  13  X  12  +  7  =  163  pennyweights;  163  -h  20 
=  8  ounces  and  3  pennyweights  remaining.  Then,  5  X  12 
-)-  8  =  68  ounces;  68  -4-  12  =  5  pounds  and  8  ounces  remain- 
ing. Then,  7  X  12  -f-  5  =  89  pounds.  The  entire  product  is 
89  pounds  8  ounces  3  pennyweights  12  grains. 

58.  Rule. — Multiply  the  number  representing  each  de- 
nomination by  the  multiplier,  and  reduce  each  product  to  the 
next  higher  denomination,  writing  the  remainders  under  each 
denomination,  and  carry  the  quotient  to  the  next,  as  in  addition 
of  denominate  numbers. 

59.  NOTE. — In  multiplication  and  division  of  denominate  num- 
bers,   it  is  sometimes  easier    to   reduce   the   number  to  the  lowest 
denomination  given  before  multiplying  or  dividing,  especially  if  the 


§  2  ARITHMETIC.  21 

multiplier  or  divisor  is  a  decimal.  Thus,  in  the  above  example,  had 
the  multiplier  been  1.2,  the  easiest  way  to  multiply  would  have  been  to 
reduce  the  number  to  grains;  then,  multiply  by  1.2,  and  reduce  the 
product  to  higher  denominations.  For  example,  7  Ib.  5  oz.  13  pwt. 
15  gr.  =  43,047  gr.  43,047  X  1.2  =  51,656.4  gr.  =  8  Ib.  11  oz.  12  pwt. 
8.4  gr.  Also,  43,047  X  12  =  516,564  gr.  =  89  Ib.  8  oz.  3  pwt.  12  gr.,  as 
before.  The  student  may  use  either  method. 


EXAMPLES  FOR  PRACTICE. 

6O.      Multiply: 

(a)  15  cwt.  90  Ib.  by  5;  (6)  12  yr.  11  mo.  3  da.  by  14;  (c)  11  mi.  145  rd. 
by  20;  (d)  12  gal.  4  pt.  by  9;  (e)  8  cd.  76  cu.  ft.  by  15;  (/)  4  hhd.  3  gal. 
1  qt.  1  pt.  by  12. 

(a)  79  cwt.  50  Ib. 

(b)  180  yr.  11  mo.  2  wk. 


Ans. 


(c)  229  mi.  20  rd. 

(d)  112  gal.  2  qt. 

(e)  128  cd.  116  cu.  ft. 
(/)  48  hhd.  40  gal.  2  qt. 


DIVISION  OF  DENOMINATE  NUMBERS. 

61.     EXAMPLE  1.— Divide  48  Ib.  11  oz.  6  pwt.  by  8. 

SOLUTION. —  Ib.          oz.        pwt.       gr. 

8  )  48  11  6  0 


6  Ib.        1  oz.      8  pwt.  6  gr.     Ans. 

EXPLANATION. — After  placing  the  quantities  as  above, 
proceed  as  follows:  8  is  contained  in  48  6  times  without 
a  remainder.  8  is  contained  in  11  ounces  once  with 
3  ounces  remaining.  3  X  20  —  60 ;  60  -f-  6  =  66  pennyweights ; 
66  pennyweights  -4-8  =  8  pennyweights  and  2  pennyweights 
remaining  ;  2  X  24  grains  =  48  grains  ;  48  grains  -4-  8 
=  6  grains.  Therefore,  the  entire  quotient  is  6  pounds 
1  ounce  8  pennyweights  6  grains. 

EXAMPLE  2.— A  silversmith  melted  up  2  Ib.  8  oz.  10  pwt.  of  silver, 
which  he  made  into  6  spoons ;  what  was  the  weight  of  each  spoon  ? 

SOLUTION. —  Ib.        oz.        pwt. 

6  )  2  8  10 

5  oz.        8  pwt.     8  gr.     Ans. 

EXPLANATION. — Since  we  cannot  divide  2  pounds  by  6,  we 
reduce  it  to  ounces.  2  pounds  =  24  ounces,  and  24  ounces 
+  8  ounces  =  32  ounces  ;  32  ounces  -4-6  =  5  ounces  and 


22  ARITHMETIC.  §  2 

2  ounces  over.  2  ounces  =  40  pennyweights.  40  penny- 
weights ~f-  10  pennyweights  =  50  pennyweights,  and  50  penny- 
weights -r-  6  =  8  pennyweights  and  2  pennyweights  over. 
2  pennyweights  =  48  grains,  and  48  grains  -4-6  =  8  grains. 
Hence,  each  spoon  contains  5  ounces  8  pennyweights 
8  grains. 

62.     EXAMPLE.— Divide  820  rd.  4  yd.  2  ft.  by  112. 
SOLUTION.—  rd.      yd.    ft.  rd.     yd.    ft.     in. 

112)820  4   2(7    1   2  5.143  Ans. 
784 

3  6  rd.  rem. 
5.5 

180 

180 

1  9  8.0  yd. 

4  yd. 

1  1  2  )  2  0  2  yd.  (  1  yd. 
112 

9  0  yd.  rem. 
3 

2  7  0  ft. 

2ft. 

112)  2~7i  ft.  (  2  ft. 
224 

4  8  ft.  rem. 
1  2 


48 

1  1  2  )  5  7  6.0  0  0  0  in.  (  5.1  4  2  8+  in.,  or  5.1  4  3  in. 
560 


480 
448 

320 

224 

960 

896 

64 


§  2  ARITHMETIC.  23 

EXPLANATION. — The  first  quotient  is  7  rods  with  36  rods 
remaining.  5.5  X  36  =  198  yards;  198  yards  +  4  yards 
=  202  yards;  202  yards  -T- 112  =  1  yard  and  90  yards  remain- 
ing. 90  X  3  =  270  feet ;  270  feet  +  2  feet  =  272  feet ;  272  feet 
-f-  112  —  2  feet  and  48  feet  remaining;  48  X  12  =  576  inches; 
576  inches  -h  112  =  5.143  inches,  nearly. 

The  preceding  example  is  solved  by  long  division,  because 
the  numbers  are  too  large  to  deal  with  mentally.  Instead  of 
expressing  the  last  result  as  a  decimal,  it  might  have  been 
expressed  as  a  common  fraction.  Thus,  576  -T-  112  =  5TW 
=  5|  inches.  The  chief  advantage  of  using  a  common  frac- 
tion is  that  if  the  quotient  be  multiplied  by  the  divisor,  the 
result  will  always  be  the  same  as  the  original  dividend. 

63.  Rule. — Find  how  many  times  the  divisor  is  contained 
in  the  first  or  highest  denomination  of  the  dividend.  Reduce 
the  remainder  (if  any]  to  the  next  lower  denomination  and 
add  to  it  the  number  in  the  given  dividend  expressing  that 
denomination.  Divide  this  new  dividend  by  the  divisor. 
The  quotient  will  be  the  next  denomination  in  the  quotient 
required.  Continue  in  this  manner  until  the  lowest  denomi- 
nation is  reached.  The  successive  quotients  will  constitute  the 
entire  quotient. 


EXAMPLES  FOB  PRACTICE. 
64.     Divide: 

(a)  376  mi.  276  rd.  by  22;  (b)  1,137  bu.  3  pk.  4  qt.  1  pt.  by  10; 
(c)  84  cwt.  48  Ib.  49  oz.  by  16;  (d)  78  sq.  yd.  18  sq.  ft.  41  sq.  in.  by  18; 
(<?)  148  mi.  64  rd.  24  yd.  by  12;  (/)  100  T.  16  cwt.  18  Ib.  11  oz.  by  15; 
(g)  36  Ib.  18  oz.  18  pwt.  14  gr.  by  8;  (k)  112  mi.  48  rd.  by  100. 

Ans.  (a)  17  mi.  41T7T  rd. ;  (d)  113  bu.  3  pk.  1  qt.  }  pt. ;  (c)  5  cwt. 
28  Ib.  3TV  oz. ;  (d)  4  sq.  rd.  4  sq.  ft.  2TB?  sq.  in. ;  (e)  12  mi.  112  rd.  2  yd*; 
(/)  6  T.  14  cwt.  41  Ib.  m  oz. ;  (g)  4  Ib.  8  oz.  7  pwt.  7f  gr.;  (k)  1  mi. 
38|f  rd. 

1.  On  Monday,  1  T.  3  cwt.  of  coal  are  burned  under  a  boiler;  on 
Tuesday,  1  T.  1  cwt.  54  Ib. ;  on  Wednesday,  1  T.  2  cwt.  16  Ib. ;  on 
Thursday,  1  T.  2  cwt.  70  Ib. ;  on  Friday,  1  T.  3  cwt.  43  Ib. ;  on  Satur- 
day, 15  cwt.  68  Ib.  How  much  coal  was  burned  during  the  week  ? 

Ans.  6  T.  8  cwt.  51  Ib. 


24  ARITHMETIC.  §  2 

2.  504  gal.  2  qt.  of  water  are  drawn  from  a  tank  containing  30  hhd. 
4  gal.  3  qt.  of  water.     How  much  water  remains  ? 

Ans.  22  hhd.  4  gal.  1  qt. 

3.  A  main  line  shaft  is  made  up  of  5  lengths,  as  follows:  16  ft.  3  in., 
15  ft.  8  in.,  15  ft.  2  in.,  14  ft.  6  in.,  12  ft.  10  in.     If  10  hangers  are  used, 
one  being  at  each  end  of  the  shaft,  what  is  the  distance  between  them, 
supposing  them  to  be  spaced  equally  ?  Ans.  8  ft.  3|  in. 

4.  If  the  distance  around  a  flywheel  is  '47  ft.  3  in.  and  the  belt 
extends  f  of  the  way  around,  what  is  the  distance  covered  by  the  belt  ? 

Ans.  28  ft.  4£  in. 

5.  A  boiler  shell  is  made  up  of  three  sheets,  each  5  ft.  6^  in.  long. 
If  the  lap  at  each  of  the  two  middle  seams  is  2|  in.,  what  is  the  length 
of  the  shell  ?  Ans.  16  ft.  3TV  in. 

6.  In  a  return-tubular  boiler  the  heating  surface  is  divided  as  fol- 
lows :    outside  of  shell,  98  sq.  ft.  9.8  sq.  in.;   heads,  5  sq.  ft.  4|  sq.  in. ; 
tubes.  683  sq.  ft.  10.75  sq.  in.     What  is  the  total  area  of  the  heating  sur- 
face in  square  feet  and  square  inches  ?         Ans.  786  sq.  ft.  25.05  sq.  in. 


ARITHMETIC. 

(PART  5.) 


1.  If  a  product  consists  of  equal  factors,  it  is  called  a 
power  of  one  of  those  equal  factors,  and  one  of  the  equal 
factors  is  called  a  root  of  the  product.     The  power  and  the 
root  are  named  according  to  the  number  of  equal  factors  in 
the   product.     Thus,  3x3,  or  9,  is  the  second  power,    or 
square,  of  3;  3  X  3  X  3,  or  27,  is  the  third  power,  or  cube, 
of  3;  3  X  3  X  3  X  3,  or  81,  is  the  fourth  power  of  3.     Also, 
3  is  the  second  root,  or  square  root,  of  9  ;  3  is  the  third 
root,  or  cube  root,  of  27  ;  3  is  the  fourth  root  of  81. 

2.  For  the  sake  of  brevity, 

3  X  3  is  written  3s,  and   read  three  square, 

or  three  exponent  two; 
3  X  3  X  3  is  written  3s,  and  read  three  cube, 

or  three  exponent  three; 
3x3x3x3  is  written  34,  and  read  three  fourth, 

or  three  exponent  four  ; 
and  so  on. 

A  number  written  above  and  to  the  right  of  another  num- 
ber, to  show  how  often  the  latter  number  is  used  as  a  factor, 
is  called  an  exponent.  Thus,  in  3",  the  number  lf  is  the 
exponent,  and  shows  that  3  is  to  be  used  as  a  factor  twelve 
times;  so  that  312  is  a  contraction  for 

3X3X3X3X3X3X3X3X3X3X3X3. 

§2 

For  notice  of  copyright,  see  page  immediately  following  the  title  page. 


26  ARITHMETIC.  §  2 

In  an  expression  like  36,  the  exponent  *  shows  how  often 
3  is  used  as  a  factor.  Hence,  if  the  exponent  of  a  number 
is  unity,  the  number  is  used  once  as  a  factor;  thus,  31  =  3, 
41  =  4,  51  =  5. 

3.  If  the  side  of  a  square  contains  5  inches,  the  area  of 
the  square  contains  5  X  5,  or  5a,  square  inches.     If  the  edge 
of  a  cube  contains  5  inches,  the  volume  of  the  cube  contains 
5  X  5  X  5,  or  5",   cubic  inches.     It  is  for  this  reason  that 
5*  and  5s  are  called  the  square  and  cube  of  5,  respectively. 

4.  To  flnd  any  power  of  a  number : 

EXAMPLE  1.— What  is  the  third  power,  or  cube,  of  35  ? 
SOLUTION.—  35  x  35  x  35, 

or  35 

3_5 

175 
105 
1225 
35 


6125 
3675 
cube  =  42875    Ans. 


EXAMPLE  2.— What  is  the  fourth  power  of  15  ? 
SOLUTION.—  15  X  15  X  15  X  15, 

or     15 
15 
75 
15 


15 


1125 

225 

3375 


16875 
3375 


fourth  power  =  50625    Ans. 


§  2 


ARITHMETIC. 


27 


EXAMPLE  3.—    1.2s  =  what  ? 
SOLUTION.—  1.2x1.2x1.2, 

or         1.2 
1.2 
1.44 
1.2 
288 
144 
cube  =  1.7  2  8     Ans. 

EXAMPLE  4.  —  What  is  the  third  power,  or  cube,  of  f  ? 


An, 


5.  Rule.  —  I.  To  raise  a  whole  number  or  a  decimal  to  any 
power,  use  it  as  a  factor  as  many  times  as  there  are  units  in 
the  exponent, 

II.  To  raise  a  fraction  to  any  power,  raise  both  the  numer- 
ator and  denominator  to  the  power  indicated  by  the  exponent. 


EXAMPLES  FOR  PRACTICE. 

6.     Raise  the  following  to  the  powers  indicated: 


(a)  85*. 

0)  Of)9- 

(c)  6.5». 

(<0  144- 

w  a)'. 

</)  (i)3. 

Or)  (I)3- 

(A)  1.4*. 


(a)  7,225. 

(*)  ttf 

(f)  42.25. 

(</)  38,416. 

(')  H- 

(/)  Mi 

(f)  si^- 

(A)  5.37824 


ETOLUTIO^. 

7.  Evolution   is   the  reverse  of   involution.     It   is  tha 
process  of  finding  the  root  of  a  number  which  is  considered 
as  a  power. 

8.  The  square  root  of  a  number  is  that  number  which, 
when  used  twice  as  a  factor,  produces  the  number. 

Thus,  2  is  the  square  root  of  4,  since  2  X  2,  or  2*  =  4. 


28  ARITHMETIC.  §  2 

9.  The  cube  root  of  a  number  is  that  number  which, 
when  used  three  times  as  a  factor,  produces  the  number. 

Thus,  3  is  the  cube  root  of  27,  since  3  X  3  X  3,  or  33  =  27. 

10.  The  radical  sign  |/,  when  placed  before  a  number, 
indicates  that  some  root  of   that  number  is  to  be  found. 
The  vinculum  is  almost  always  used  in  connection  with  the 
radical  sign,  as  shown  in  Art.  11. 

11.  The  Index  of  the  root  is  a  small  figure  placed  over 
and  to  the  left  of  the  radical  sign,  to  show  what  root  is  to 
be  found. 

Thus,  1/100  denotes  the  square  root  of  100. 
I/T25  denotes  the  cube  root  of  125. 
4/256  denotes  the  fourth  root  of  256,  and  so  on. 

12.  When  the  square  root  is  to  be  extracted,  the  index 
is  generally  omitted.     Thus,  4/100  indicates  the  square  root 
of  100.     Also,  |/225  indicates  the  square  root  of  225. 


SQUARE    ROOT. 

13.  The  largest  number  that  can  be  written  with  one 
figure  is  9,  and  9s  =  81 ;  the  largest  number  that  can  be 
written  with  two  figures  is  99,  and  99"  =  9,801;  with  three 
figures  999,  and  999"  =  998,001;  with/owr  figures  9,999,  and 
9,999'=  99,980,001,  etc. 

In  cadi  of  the  above  it  will  be  noticed  that  the  square  of 
the  number  contains  just  twice  as  many  figures  as  the 
number. 

In  order  to  find  the  square  root  of  a  number,  the  first 
step  is  to  find  how  many  figures  there  will  be  in  the  root. 
This  is  done  by  pointing  off  the  number  into  periods  of  two 
figures  each,  beginning  at  the  decimal  point.  The  number 
of  periods  will  indicate  the  number  of  figures  in  the  root. 

Thus,  the  square  root  of  83,740,801  must  contain  4  figures, 
since,  pointing  off  the  periods,  we  get  83'74'08'01,  or  4  periods ; 
consequently,  there  must  be  4  figures  in  the  root.  In  like 
manner,  the  square  root  of  50,625  must  contain  3  figures, 


§  2  ARITHMETIC.  29 

since  there  are  (5'06'25)  3  periods.  The  extreme  left-hand 
period  may  contain  either  1  or  2  figures,  according  to  the 
size  of  the  number  squared. 

14.     EXAMPLE.— Find  the  square  root  of  31,505,769. 

root 

SOLUTION.—       (a)         5  3  1'5  0'5  7'6  9  ( 5  6  1  3    Ans. 

5  (&)      2  5 

(d)     1  0  0  (c)         650 

6  636 

rb~6  (e) 


1120 
1 


1121 

1 

11220 


11223 

EXPLANATION. — Pointing  off  into  periods  of  two  figures 
each,  it  is  seen  that  there  are  4  figures  in  the  root.  Now, 
find  the  largest  single  number  whose  square  is  less  than 
or  equal  to  31,  the  first  period.  This  is  evidently  5,  since 
62  =  36,  which  is  greater  than  31.  Write  it  to  the  right, 
as  in  long  division,  and  also  to  the  left,  as  shown  at  (a). 
This  is  the  first  figure  of  the  root.  Now,  multiply  the 
5  at  (a)  by  the  5  in  the  root,  and  write  the  result  under  the 
first  period,  as  shown  at  (b].  Subtract  and  obtain  6  as  a 
remainder. 

Bring  down  the  next  period,  50,  and  annex  it  to  the 
remainder  6,  as  shown  at  (c),  which  we  call  the  dividend. 
Add  the  root  already  found  to  the  5  at  (a),  getting  10,  and 
annex  a  cipher  to  this  10,  thus  making  it  100,  as  shown  at 
(d),  which  we  call  the  trial  divisor.  Divide  the  divi- 
dend (c)  by  the  trial  divisor  (d]  and  obtain  6,  which  is 
probably  the  next  figure  of  the  root.  Write  6  in  the  root, 
as  shown,  and  also  add  it  to  100,  the  trial  divisor,  making  it 
106.  This  is  called  the  complete  divisor. 


30  ARITHMETIC.  §  2 

Multiply  this  by  6,  the  second  figure  in  the  root,  and  sub- 
tract the  result  from  the  dividend  (c).  The  remainder  is 
14,  to  which  annex  the  next  period,  making  it  1457,  as 
shown  at  (e),  which  we  call  the  new  dividend.  Add  the 
second  figure  of  the  root  to  the  complete  divisor,  106,  and 
annex  a  cipher,  thus  getting  1120.  Dividing  1457  by  1120, 
we  get  1  as  the  next  figure  of  the  root.  Adding  this  last 
figure  of  the  root  to  1120,  multiplying  the  result  by  it,  and 
subtracting  from  1457,  the  remainder  is  336. 

Annexing  the  next  and  last  period,  69,  the  result  is  33669. 
Now,  adding  the  last  figure  of  the  root  to  1121  and  annex- 
ing a  cipher  as  before,  the  result  is  11220.  Dividing 
33669  by  11220,  the  result  is  3,  the  fourth  figure  in  the 
root.  Adding  it  to  11220  and  multiplying  the  sum  by  it, 
the  result  is  33669.  Subtracting,  there  is  no  remainder; 
hence,  |/31, 505,769  =  5,613. 

15.  The  square  of  any  number  wholly  decimal  always 
contains   twice  as    many  figures  as  the   number  squared. 
For  example,  .I3  =  .01,  .13'  =  .0169,  .751s  =  .564001,  etc. 

16.  It  will  also  be  noticed  that  the  number  squared  is 
always  less  than  the  decimal.     Hence,  the  square  root  of  a 
number  wholly  decimal  is  greater  than  the  number  itself. 
If  it  be  required  to  find  the  square  root  of  a  decimal,  and 
the  decimal  has  not  an  even  number  of  figures  in  it,  annex 
a  cipher.     The  best  way  to  determine  the  number  of  figures 
in  the  root  of  a  decimal-  is  to  begin  at  the  decimal  point,  and, 
going  towards  the  right,  point  off  the  decimal  into  periods 
of  two  figures  each.     Then,  if  the  last  period  contains  but 
one  figure,  annex  a  cipher. 

17.  EXAMPLE.— What  is  the  square  root  of  .000576  ? 

root 

SOLUTION.—  2  .0  O'O  5'7  6  (  .0  2  4    Ans. 

2  4 

40  176 

4  176 

"44 


§  2  ARITHMETIC.  31 

EXPLANATION. — Beginning  at  the  decimal  point  and  point- 
ing off  the  number  into  periods  of  'two  figures  each,  it  is 
seen  that  the  first  period  is  composed  of  ciphers;  hence,  the 
first  figure  of  the  root  must  be  a  cipher.  The  remaining 
portion  of  the  solution  should  be  perfectly  clear  from  what 
has  preceded. 

18.  If  the  number  is  not  a  perfect  power,  the  root  will 
consist  of  an  interminable  number  of  decimal  places.     The 
result  may  be  carried  to  any  required  number  of  decimal 
places  by  annexing  periods  of  two  ciphers  each  to  the  num- 
ber. 

19.  EXAMPLE. — What  is  the  square  root  of  3  ?     Find  the  result 
to  five  decimal  places. 

root 

SOLUTION.—    1  3.0  O'O  O'O  O'O  O'O  0  (  1.7  3  2  0  5+    Ans. 


20  200 

7  189 

¥7  1100 

7  1029 


340  7100 

3  6924 


343  1760000 

3  1732025 


3460  27975 

2 


3462 
2 

346400 
5 

346405 

EXPLANATION. — Annexing  5  periods  of  two  ciphers  each, 
to  the  right  of  the  decimal  point,  the  first  figure  of  the  root 
is  1.  To  get  the  second  figure,  we  find  that,  in  dividing 
200  by  20,  it  is  10.  This  is  evidently  too  large. 

Trying  9,  we  add  9  to  20  and.  multiply  29  by  9 ;  the  result 
is  261?  a  result  which  is  considerably  larger  than  200;  hence? 


32  ARITHMETIC.  §  2 

9  is  too  large.  In  the  same  way,  it  is  found  that  8  is  also  too 
large.  Trying  7,  7  times  27  are  189,  a  result  smaller  than 
200 ;  therefore,  7  is  the  second  figure  of  the  root.  The  next 
two  figures,  3  and  2,  are  easily  found.  The  fifth  figure  in 
the  root  is  a  cipher,  since  the  trial  divisor,  34640,  is  greater 
than  the  new  dividend,  17600.  In  a  case  of  this  kind  we 
annex  another  cipher  to  34640,  thereby  making  it  346400, 
and  bring  down  the  next  period,  making  the  17600,  1760000. 
The  next  figure  of  the  root  is  5,  and,  as  we  now  have  five 
decimal  places,  we  will  stop. 

The  square  root  of  3  is,  then,  1.73205+. 

2O.     EXAMPLE. — What  is  the  square  root   of  .3  to  five  decimal 
places  ? 

SOLUTION. — 


5 
5 

loo 

4 

104 

4 

1080 

7 

1087 

7 

10940 

7 

10947 

7 

109540 


.3  O'O  O'O  O'O  O'O  0  (  .5  4  7  7  2+  Ans. 
25 


416 

8400 
7609 


79100 
76629 

247100 
21  9084 


109542 

EXPLANATION. — In  the  above  example  we  annex  a  cipher 
to  .3,  making  the  first  period  .30,  since  every  period  of  a 
decimal,  as  was  mentioned  before,  must  have  two  figures  in 
it.  The  remainder  of  the  work  should  be  perfectly  clear. 

21.  If  it  is  required  to  find  the  square  root  of  a  mixed 
number,  begin  at  the  decimal  point  and  point  off  the  periods 


§  2  ARITHMETIC.  33 

both  to  the  right  and  to  the  left.  The  manner  of  finding 
the  root  will  then  be  exactly  the  same  as  in  the  previous 
cases. 

22.  EXAMPLE.— What  is  the  square  root  of  258.2449  ? 

root 

SOLUTION.—  1  2'5  8.2  4'4  9  ( 1  6.0  7    Ans. 

1  1 

20  158 

6  156 

¥e  22449 

6  22449 

3200 

7 

3207 

EXPLANATION. — In  the  above  example,  since  320  is  greater 
than  224,  we  place  a  cipher  for  the  third  figure  of  the  root 
and  annex  a  cipher  to  320,  making  it  3200.  Then,  bringing 
down  the  next  period,  49,  7  is  found  to  be  the  fourth  figure 
of  the  root.  Since  there  is  no  remainder,  the  square  root 
of  258.2449  is  16.07. 

23.  Proof.  —  To  prove  square  root,    square   the  result 
obtained.     If  the  number  is  an  exact  power,  the  square  of  the 
root  will  equal  it;  if  it  is  not  an  exact  power,  the  square  of 
the  root  will  very  nearly  equal  it. 

24.  Rule. — I.     Begin  at   units  place  and  separate   the 
number  into  periods  of  tivo  figures  each,  proceeding  from  left 
to  riglit  witJi  t/ie  decimal  part,  if  there  be  any. 

II.  Find  the  greatest  number  whose  square  is  contained  in 
the  first,  or  left-hand,  period.     Write  this  number  as  the  first 
figure  in  the  root;  also,  write  it  at  the  left  of  the  given 
number. 

Multiply  this  number  at  the  left  by  the  first  figure  of  th$ 
root,  and  subtract  the  result  from  the  first  period;  then, 
annex  the  second  period  to  the  remainder. 

III.  Add  the  first  figure  of  the  root  to  the  number  in  the 
first  cohimn  on  the  left  and  annex  a  cipher  to  the  result; 
this  is  the  trial  divisor.     Divide  the  dividend  by  the  trial 

II.  8.    I.—7 


34  ARITHMETIC.  §  2 

divisor  for  the  second  figure  in  the  root  and  add  this  figure 
to  the  trial  divisor  to  form  the  complete  divisor.  Multiply 
the  complete  divisor  by  the  second  figure  in  the  root  and  sub- 
tract this  result  from  the  dividend.  (//"  this  result  is  larger 
than  the  dividend,  a  smaller  number  must  be  tried  for  the 
second  figure  of  the  root.}  Now  bring  down  the  third  period 
and  annex  it  to  the  last  remainder  for  a  new  dividend.  Add 
the  second  figure  of  the  root  to  the  complete  divisor  and  annex 
a  cipher  for  a  new  trial  divisor. 

IV.  Continue   in   this   manner  to  the  last  period,  after 
which,  if  any  additional  places  in  the  root  are  required,  bring 
down  cipher  periods  and  continue  the  operation. 

V.  If  at  any  time  the  trial  divisor  is  not  contained  in  the 
dividend,  place  a  cipher  in  the  root,  annex  a  cipher  to  the  trial 
divisor,  and  bring  down  another  period. 

VI.  If  the  root  contains  an  interminable  decimal  and  it 
is  desired  to  terminate  the  operation  at  some  point,  say,  the 
fourth  decimal  place,  carry  tlie  operation  one  place  farther, 
and  if  the  fifth  figure  is  5  or  greater,  increase  the  fourth 
figure  by  1  and  omit  the  sign  -J-. 

25.  Short  Method. — If  the  number  whose  root  is  to  be 
extracted  is  not  an  exact  square,  the  root  will  be  an  inter- 
minable decimal.  It  is  then  usual  to  extract  the  root  to  a 
certain  number  of  decimal  places.  In  such  cases,  the  work 
may  be  greatly  shortened  as  follows:  Determine  to  how 
many  decimal  places  the  work  is  to  be  carried,  say  5,  for 
example ;  add  to  this  the  number  of  places  in  the  integral 
part  of  the  root,  say  2,  for  example,  thus  determining  the 
number  of  figures  in  the  root,  in  this  case  5  +  2  —  7.  Divide 
this  number  by  2  and  take  the  next  higher  number.  In  the 
above  case,  we  have  7  -f-  2  =  3£;  hence,  we  take  4,  the  next 
higher  number.  Now  extract  the  root  in  the  usual  manner 
until  the  same  number  of  figures  have  been  obtained  as  was 
expressed  by  the  number  obtained  above,  in  this  case  4. 
Then  form  the  trial  divisor  in  the  usual  manner,  but  omit- 
ting to  annex  the  cipher;  divide  the  last  remainder  by  the 
frial  divisor,  as  in  long  division,  obtaining  as  many  figures 


ARITHMETIC. 


35 


of  the  quotient  as  there  are  remaining  figures  of  the  root, 
in  this  case  7  —  4  =  3.  The  quotient  so  obtained  is  the 
remaining  figures  of  the  root. 

Consider  the  example  in  Art.  2O.  Here  there  are  5  figures 
in  the  root.  We  therefore  extract  the  root  to  3  figures  in  the 
usual  manner,  obtaining  .547  for  the  first  three  root  figures. 
The  next  trial  divisor  is  1094  (with  the  cipher  omitted)  and 
the  last  remainder  is  791.  Then,  791  -+  1094  =  .723,  and  the 
next  two  figures  of  the  root  are  72,  the  whole  root  being 
.54772+-  Always  carry  the  division  one  place  farther  than 
desired,  and  if  the  last  figure  is  5  or  greater,  increase  the 
preceding  figure  by  1.  This  method  should  not  be  used 
unless  the  root  contains  five  or  more  figures. 

NOTE. — If  the  last  figure  of  the  root  found  in  the  regular  manner 
is  a  cipher,  carry  the  process  one  place  farther  before  dividing  as 
described  above. 


EXAMPLES  FOR  PRACTICE. 

26.  Find  the  square  root  of  : 

(a)  186,624. 

(b)  2,050,624. 

(c)  29,855,296. 

(d)  .0116964. 
(<?)  198.1369. 

(/)  994,009.  Ans. 

(g)  2.375  to  four  decimal  places. 

(ft)  1.625  to  three  decimal  places. 

(/)  .3025. 

(/)  .571428. 

(k)  .78125. 


(*) 
(d) 


432. 

1,432. 

5,464. 

.1081+. 

14.0761. 

997. 

1.5411. 

1.275. 

.55. 

.7559+. 


CUBE   ROOT. 

27.  By  a  method  similar  to  the  foregoing  for  extracting 
square  root,  the  cube  root  of  any  number  may  be  found. 
The  labor  involved  in  finding  the  cube  root  of  a  number  is 
so  great  and  the  necessity  of  extracting  the  cube  root  so 
seldom  arises  that  we  have  deemed  it  best  to  omit  the  exact 
method  and  substitute  instead  an  approximate  method, 


36  ARITHMETIC.  §  2 

which  will  answer  the  student's  purposes  fully  as  well  as  the 
exact  method,  and  is  far  easier  to  learn  and  apply. 

28.  In  applying  this  approximate   method,  we   find  the 
first  three  or  four  figures  of  the  root  by  means  of  a  table  in 
a  manner  to  be  described  presently  and  then  find  more  figures, 
if  necessary,  as  described  in  Arts.  35  and  36. 

29.  In  any  number,  the  figures  beginning  with  the  first 
digit*  at  the  left  and  ending  with  the  last  digit  at  the  right 
are  called  the  significant  figures  of  the  number.     Thus, 
the  number  405,800  has  the  four  significant  figures  4,  0,  5,  8; 
and  the  number  .000090067  has  the   five  significant  figures 
9.  0,  0,  6,  and  7. 

The  part  of  a  number  consisting  of  its  significant  figures 
is  called  the  significant  part  of  the  number.  Thus,  in  the 
number  28,070,  the  significant  part  is  2807;  in  the  number 
.00812,  the  significant  part  is  812;  and  in  the  number  170.3, 
the  significant  part  is  1703. 

In  speaking  of  the  significant  figures  or  of  the  significant 
part  of  a  number,  we  consider  the  figures,  in  their  proper 
order,  from  the  first  digit  at  the  left  to  the  last  digit  at  the 
right,  but  we  pay  no  attention  to  the  position  of  the  decimal 
point.  Hence,  all  numbers  that  differ  only  in  the  position  of 
the  decimal  point  have  the  same  significant  part.  For 
example,  .002103,  21.03,  21,030,  and  210,300  have  the  same 
significant  figures  2,  1,  0,  and  3,  and  the  same  significant 
part  2103. 

30.  In  the  same  manner  as  in  the  case  of  square  root,  it 
can    be    shown    that  the   periods   into  which    a  number   is 
divided  whose  cube  root  is  to  be  extracted,  must  contain 
three  figures,   except  that  the  first  (left-hand)  period  of  a 
whole  or  mixed  number   may  contain    one,   two,  or  three 
figures.     Hence,  the  first  step  in  extracting  the  cube  root  is 
to  point  off  the  number  into  periods  of  three  figures  each, 
beginning  at  the  decimal  point  and  proceeding  to  the  left  and 
to  the  right.     The  remaining  steps  are  best  illustrated  by  an 
example. 

*  A  cipher  is  not  a  digit. 


ARITHMETIC. 


37 


SQUARES  AND  CUBES. 


X... 

Square. 

Cube. 

No. 

Square. 

Cube. 

No. 

Square. 

Cube. 

1 

1 

1 

34 

11'56 

39'304 

67 

44'89 

300'  763 

2 

4 

8 

35 

12'25 

42'875 

68 

46'24 

314'432 

3 

9 

27    • 

36 

12'96 

46'656 

69 

47'61 

328'509 

4 

16 

64 

37 

13'69 

50'653 

70 

49'00 

343'000 

5 

25 

125 

38 

14'44 

54'872 

71 

50'41 

357'911 

6 

36 

216 

39 

15'21 

59'319 

72 

51'84 

373'248 

7 

49 

343 

40 

16'00 

64'000 

73 

53'29 

389'017 

8 

64 

512 

41 

16'81 

68'921 

74 

54'76 

405'224 

9 

81 

729 

42 

17'64 

74'088 

75 

56'25 

421  '875 

10 

I'OO 

I'OOO 

43 

18'49 

79'507 

76 

57'76 

438'976 

11 

T21 

1'331 

44 

19'36 

85184 

77 

59'29 

456'  533 

12 

1'44 

1'728 

45 

20'25 

91125 

78 

60'84 

474'  552 

18 

1'69 

2'197 

46 

2116 

97'336 

79 

62'41 

493'039 

14 

1'96 

2'744 

47 

22'09 

103'823 

80 

64'00 

512'000 

15 

2'25 

3'375 

48 

23'04 

110'592 

81 

65'61 

531  '441 

16 

2'56 

4'096 

49 

24'01 

11  7'  649 

82 

67'24 

551'368 

17 

2'89 

4'913 

50 

25'00 

125'000 

83 

68'89 

571'787 

18 

324 

5'832 

51 

26'01 

132'651 

84 

70'56 

592'  704 

19 

3'61 

6'859 

52 

27'04 

140'608 

85 

72'25 

614125 

20 

4'00 

8'000 

53 

28'09 

148'877 

86 

73'96 

636'056 

21 

4'41 

9'261 

54 

2916 

157'464 

87 

75'69 

658'503 

22 

4'84 

10'648 

55 

30'25 

166'375 

88 

77'44 

681'472 

23 

5'29 

12167 

56 

31'36 

175'616 

89 

79'21 

704'969 

24 

5'76 

13'824 

57 

32'49 

185193 

90 

Sl'OO 

729'000 

25 

6'25 

15'625 

58 

33'64 

195112 

91 

82'81 

753'571 

26 

6'76 

17'576 

59 

34'81 

205'379 

92 

84'  64 

778'688 

27 

7'29 

19'683 

60 

36'00 

216'000 

93 

86'49 

804'  357 

28 

7  '84 

21'952 

61 

37'21 

226'981 

94 

88'36 

830'584 

29 

8'41 

24'  389 

62 

38'44 

238328 

95 

90'25 

857'375 

30 

9'00 

27'000 

63 

39'69 

250'047 

96 

9216 

884'  736 

31 

9'61 

29'791 

64 

40'96 

262144 

97 

94'09 

912'673 

32 

1024 

32768 

65 

42'25 

274'625 

98 

96'04 

941192 

33 

10'89 

35'937 

66 

43'56 

287'496 

99 

98'01 

970'299 

EXAMPLE.— Extract  the  cube  root  of  160,524. 

SOLUTION. — Pointing  off  into  periods  of  three  figures  each,  we 
have  160'524,  two  periods  of  three  figures  each;  hence,  the  whole- 
number  part  of  the  root  will  contain  two  figures.  Referring  to  the 
table  of  Squares  and  Cubes,  and  looking  in  the  columns  headed 
"Cube"  (these  columns  contain  the  cubes  of  the  numbers  in  the 


38  ARITHMETIC.  §  2 

columns  headed  "No."),  we  see  that  the  given  number  falls  between 
157,464,  the  cube  of  54,  and  166,375,  the  cube  of  55.  Hence,  the  required 
cube  root  is  54+.  Now  find  the  difference  between  the  two  numbers  in 
the  table  between  which  the  given  number  falls,  and  call  the  result  the 
first  difference,  in  this  case  166,375  —  157,464  =  8,911.  Also  find 
the  difference  between  the  given  number  and  the  smaller  of  the  two 
numbers  in  the  table  between  which  it  falls,  and  call  the  result  the 
second  difference,  in  this  case  160,524  -  157,464  =  3,060.  Now 
divide  the  second  difference  by  the  first  difference  and  carry  the  result 
to  three  decimal  places,  increasing  the  second  figure  of  the  decimal  by  1 
if  the  third  figure  is  5  or  greater,  thus:  3,060  -*-  8,911  =  .343  +  ,  or  .34. 
The  decimal  .34  is  the  next  two  figures  of  the  root.  Hence,  the  cube 
root  of  160,524,  or  |/160,524,  is  54.34  to  four  significant  figures.  Ans. 

31.  Had  the  given  number  been  160.524,  .160524, 
.000160524,  or  160,524,000,  the  figures  of  the  root  would  have 
been  exactly  the  same,  the  position  of  the  decimal  point  only 
being  changed,  the  reason  being  that  there  are  the  same  num- 
ber of  figures  in  the  left-hand  period  of  the  significant  part 
of  the  given  number.  Hence,  4/160.524  =  5.434+,  |/.  160524 
=  .5434+,  4/.  000160524  =  .05434+,  and  1/160,524,000 
—  543.4+. 

Suppose,  however,  that  the  given  number  had  been 
16,052.4,  16.0524,  or  .0160524,  etc. ;  then,  pointing  if  off  into 
periods,  in  the  usual  manner,  we  should  have  16'052.400, 
16.052'400,  or  .016'052'400,  etc.,  and  the  left-hand  period  of 
the  significant  part  of  the  number  contains  only  two  signifi- 
cant figures,  instead  of  three,  as  in  the  former  case  ;  in  other 
words,  the  left-hand  period  is  16  instead  of  160.  Since  the 
cube  root  of  16  is  entirely  different  from  the  cube  root  of  160, 
it  is  evident  that  great  care  must  be  taken  to  point  off  the 
periods  correctly. 

EXAMPLE.—  ^16,052.4  =  ? 

SOLUTION. — Pointing  off  into  periods  of  three  figures  each,  we  have 
16'052.400,  annexing  two  ciphers  to  complete  the  right-hand  decimal 
period.  Referring  to  the  table,  we  see  that  only  two  periods  are  given ; 
we 'therefore  drop  the  third  period  and  find  the  cube  root  of  16'052. 
The  period  dropped  will  in  no  case  affect  the  root  when  only  four  figures 
are  found ;  but  had  the  period  dropped  been  500  or  larger,  1  would  have 
been  added  to  the  last  figure  of  the  second  period.  Referring  to  the 
table,  16  052  lies  between  15'625,  the  cube  of  25,  and  17'576,  the  cube 


§  2  ARITHMETIC.  39 

of  26.  The  first  difference  is  17,576  —  15,625  =  1 ,951 ;  the  second  differ- 
ence  is  16.052  -  15,625  =  427;  and  427  +- 1,951  =:  .218+,  or  .22.  Hence, 
4X16,052.4  =  25.22  to  four  figures.  Ans. 

32.  EXAMPLE.— What  is  the  cube  root  of  .003  ? 

SOLUTION. — Since  the  number  is  entirely  decimal,  the  root  is  entirely 
decimal.  The  given  number  has  but  one  period  and  but  one  figure  in 
that  period  to  be  considered,  since  the  two  ciphers  on  the  left  are  not 
significant  figures.  The  figures  of  the  root  will  be  exactly  the  same  as 
those  in  the  cube  root  of  3  or  the  cube  root  of  3,000;  hence,  for  con- 
venience, we  neglect  the  decimal  point  for  the  present  and  find  the  cube 
root  of  3,000.  Referring  to  the  table,  3'000  lies  between  2'744,  the 
cube  of  14,  and  3'375,  the  cube  of  15 ;  the  first  two  figures  of  the  root  are, 
therefore,  14.  The  first  difference  is  3,375-2,744  =  631;  the  second 
difference  is  3,000  -  2,744  =  256 ;  and  256  -H  631  =  .405+  or  .41.  There- 
fore, the  first  four  figures  of  the  root  are  1441.  To  locate  the  decimal 
point,  we  employ  the  following  principle :  The  cube  root  of  any  number 
•wholly  decimal  will  have  as  many  ciphers  between  the  decimal  and 
the  first  significant  figure  of  the  root  as  there  are  cipher  periods  of 
three  ciphers  each  between  decimal  point  and  the  first  significant 
figure  of  the  given  number.  In  the  present  example  there  are  no 
cipher  periods  between  the  decimal  point  of  the  given  number,  .003,  and 
the  first  significant  figure;  hence,  I/.  003  =  .1441.  Ans. 

33.  We  extracted  the  cube  root  of  .003  by  assuming  it  to 
be  3,000  instead  of  3,  in  order  to  obtain  two  periods  and  get 
two  figures  of  the  root  from  the  table.     The  fourth  figure  of 
the  root  just  found  is  not  quite  correct — it  ought  to  have 
been  2  instead  of  1 — in  other  words,  the  fourth  figure  can- 
not always  be  relied  upon  as  being  absolutely  exact;  it  may 
be  1  or  2  less. 

34.  Rule. — Point  off  the  number   into  periods  of  three 
figures  each,  beginning  at  the  decimal  point,  proceeding  to  the 
left  and  right.     Consider  the  first  two  periods  and  find  in  the 
table  in  the  columns  headed  "  Cube  "  the  two  numbers  between 
which  the  first  two  periods  of  the  given  number  falls.      The 
first  two  figures  of  the  root  will  be  found  opposite  the  smaller 
of  the  two  numbers  in  the  nearest  column  to  the  left,  headed 
" No."     Find  the  difference  between  the  two  numbers  in  the 
column  headed  "  Cube" ;  also  find  the  difference  between  the 
smaller  of  these  numbers  and  the  first  two  periods  of  the  given 
number.     Divide  the  second  difference  by  the  first,  carrying 


40  ARITHMETIC.  §  2 

the  quotient  to  three  figures  and  increasing  the  second  figure 
by  1,  if  the  third  figure  is  5  or  greater.  The  first  tiuo  figures 
of  this  quotient  will  be  the  third  aud fourth  figures  oftlie  root. 

35.  As   stated   in   Art.   33,    the   fourth   figure   cannot 
always  be  depended  upon;  hence,  three  figures  of  the  root 
are  all  that  we  can  be  absolutely  sure  of  as  being  correct. 
If  it  should  be  necessary  to  find  more  than  three  or  four 
figures  of  the  root,  it  can  be  easily  done  as  follows: 

Find  the  first  four  figures  as  above  described  and  locate 
the  decimal  point.  Divide  the  given  number  by  the  root  as 
found  and  carry  the  quotient  to  6  or  7  figures.  Divide  this 
quotient  by  the  root,  and  carry  the  quotient  to  6  figures. 
Then  add  the  last  quotient  and  the  two  divisors  together 
and  divide  the  sum  by  3;  the  result  will  be  the  root  correct 
to  5  or  6  figures. 

36.  EXAMPLE. — Find  the  cube  root  of  160,524  to  7  figures. 
SOLUTION.— In  Art.  3O,  the  root  to  four  figures  was  found  to  be 

54.34.  Proceeding  as  in  Art.  35,  160,524  -4-  54.34  =  2,954.067 ;  2,954.067 
-f-  54.34  =  54.3626;  54.34  +  54.34  +  54.36266  =  163.04266;  163.0426  -=-  3 
=  54.34755.  Therefore,  ^160,524  =  54.34755,  correct  to  7  figures.  Ans. 

37.  The  method  illustrated  in  the  example  of  Art.  37 
is  perfectly  general,  but   if  5  figures  only  of  the  root  are 
required,  the  fourth  and  fifth  figures  may  be  determined  in 
an  easier  manner  'by  adding  a  correction  to  the  root  as  found 
by  applying  the  rule  of  Art.  34.     The  correction  is  deter- 
mined as  follows:     When  dividing  the  second  difference  by 
the  first  difference  carry  the.  quotient  to  4  decimal  places, 
increasing  the  fourth  figure  by  1  if  the  fifth  figure  is  5  or 
greater.     Consider  the  quotient  as  a  decimal  and  subtract 
it  from  1.     Multiply  the  remainder  by  the  quotient  and 
divide  the  product  by  the  first  two  figures  of  the  root,  as 
found  from  the  table,  carrying  the  quotient  to  four  decimal 
places.     Then  add  this  quotient  to  the  quotient  previously 
found,  and  the  sum  will  be  the  third,  fourth,  and  fifth  figures 
of  the  root.    Thus,  referring  to  Art.  3O,  the  quotient  is .  3434 ; 
1  -  .3434  =  .6566;  .3434  X  .6566 -J-  54  =  .0042;  .3434  +  . 0042 
=  .3476.     The  root  is,  therefore,  54.348  to  five  figures. 


§  2  ARITHMETIC.  41 

RATIO. 

38.  Suppose  that  it  is  desired  to  compare  two  numbers, 
say  20  and  4.      If  we  wish  to  know  how  many  times  larger 
20  is  than  4,  we  divide  20  by  4  and  obtain  5  for  the  quotient; 
thus,  20  -f-  4  =  5.      Hence,  we  say  that  20  is  5  times  as  large 
as  4,  i.  e.,  20  contains  5  times  as  many  units  as '4.     Again, 
suppose  we  desire  to  know  what  part  of  20  is  4.     We  then 
divide  4  by  20  and  obtain  |;  thus,  4  -4-  20  =  |,  or  .2.     Hence, 
4  is  £,  or  .2,  of  20.     This  operation  of  comparing  two  num- 
bers is  termed  finding  the  ratio  of  the  two  numbers.     Ratio, 
then,  is  a  comparison.     It  is  evident  that  the  two  numbers 
to  be  compared  must  be  expressed   in   the  same   unit;  in 
other  words,  the  two  numbers  must  both  be  abstract  num- 
bers or  concrete  numbers  of  the  same  kind.     For  example, 
it  would  be  absurd  to  compare  20  horses  with  4  birds,  or 
20  horses  with  4.      Hence,  ratio  may  be  defined  as  a  com- 
parison between  two  numbers  of  the  same  kind. 

39.  A  ratio  may  be  expressed  in  three  ways ;  thus,  if  it  is 
desired  to  compare  20  and  4  and  express  this  comparison  as 

20 
a  ratio,  it  may  be   done  as  follows :    20  -4-  4 ;    20  :  4,  or  — . 

All  three  are  read  the  ratio  of  20  to  4.     The  ratio  of  4  to  20 

would  be  expressed  thus:    4-4-20;    4  :  20,  or—.     The  first 

ii(j 

method  of  expressing  a  ratio,  although  correct,  is  seldom  or 
never  used ;  the  second  form  is  the  one  oftenest  met  with, 
while  the  third  form,  called  the  fractional  form,  possesses 
great  advantages  to  students  of  algebra  and  of  higher 
mathematical  subjects.  The  second  form  is  better  adapted 
to  arithmetical  subjects  and  is  one  we  shall  ordinarily  adopt. 

40.  The  terms  of  a  ratio  are   the  two  numbers  to  be 
compared;  thus,  in  the  above  ratio,  20  and  4  are  the  terms. 
When  both  terms  are  considered  together,  they  are  called  a 
couplet ;  when  considered  separately,  the  first  term  is  called 
the   antecedent   and    the   second    term    the   consequent. 
Thus,  in  the  ratio  20  :  4,  20  and  4  form  a  couplet,  and  20  is 
the  antecedent  and  4  the  consequent. 


42  ARITHMETIC.  §  2 

41.  A  ratio  may  be  direct  or  inverse.  The  direct  ratio 
of  20  to  4  is  20  :  4,  while  the  inverse  ratio  of  20  to  4  is  4  :  20. 
The  direct  ratio  of  4  to  20  is  4  :  20,  and  the  inverse  ratio  is 
20  :  4.  An  inverse  ratio  is  sometimes  called  a  reciprocal 
ratio.  The  reciprocal  of  a  number  is  1  divided  by  the 
number.  Thus,  the  reciprocal  of  17  is  y^;  of  f  is  1  -h  f  =  f  ; 
i.  e.,  the  reciprocal  of  a  fraction  is  the  fraction  inverted. 
Hence,  the  inverse  ratio  of  20  to  4  may  be  expressed  as  4  :  20, 
or  as  ^  :  £.  Both  have  equal  values  ;  for,  4  -4-  20  —  -|,  and 


42.  The  term  vary  implies  a  ratio.     When  we  say  that 
two  numbers  vary  as  some  other  two  numbers,  we  mean  that 
the  ratio  between  the  first  two  numbers  is  the  same  as  the 
ratio  between  the  other  two  numbers. 

43.  The  value  of  a  ratio  is  the  result  obtained  by  per- 
forming the  division  indicated.     Thus,  the  value  of  the  ratio 
20  :  4  is  5  —  it  is  the  quotient  obtained  by  dividing  the  ante- 
cedent by  the  consequent.     The  value  of  a  ratio  is  always 
an  abstract    number,  regardless  of    whether   the  terms  are 
abstract  or  concrete  numbers 

44.  When  a  ratio  is  expressed  in  words,  as  the  ratio  of 
20  to  4,  the  first  number  named  is  always  regarded  as  the 
antecedent  and  the  second,  as  the  consequent,  without  regard 
to  whether  the  ratio  itself  is  direct  or  inverse.      When  not 
otherwise  specified,  all  ratios  are  understood  to  be  direct.     To 
express  an  inverse  ratio,  the  simplest  way  of  doing  it  is  to 
express  it  as  if  it  were  a  direct  ratio,  with  the  first  number 
named  as  the  antecedent,  and  then  transpose  the  antecedent 
to  the  place  occupied  by  the  consequent  and  the  consequent  to 
the  place  occupied  by  the  antecedent  ;  or  if  expressed  in  the 
fractional  form,  invert  the  fraction.     Thus,  to  express  the 
inverse  ratio  of  20  to  4,  first  write  it  20  :  4,  and  then,  trans- 
posing the  terms,  as  4  :  20  ;  or  as  ?T°-,  and  then  inverting,  as  2V 
Or,  the  reciprocals  of  the  numbers  may  be  taken,  as  explained 
above.     To  invert  a  ratio  is  to  transpose  its  terms. 


§  2  ARITHMETIC.  43 

45.  Instead  of  expressing  the  value  of  a  ratio  by  a  single 
number,  as  above,  it  is  convenient  to  express  it  by  means  of 
another  ratio  in  which  the  consequent  is  1.  Thus,  suppose 
that  it  is  desired  to  find  the  ratio  of  the  weights  of  two  pieces 
of  iron,  one  weighing  45  pounds  and  the  other  weighing 
30  pounds.  The  ratio  of  the  heavier  to  the  lighter  is  then 
45  :  30,  an  inconvenient  expression.  Using  the  fractional 

form,  we  have  ; — .      Dividing  both  terms  by  30,*  the  conse- 
30 

quent,  we  obtain  -^,  or  1^  :  1.     This  is  the  same  result  as 
obtained  above,  for  1£  -r-  1  =  1£,  and  45  -7-  30  =  1£. 


PROPORTION. 

46.  Proportion  is  an  equality  of  ratios,  the  equality 
being  indicated  by  the  double  colon  (::)  or  by  the  sign  of 
equality  (=).  Thus,  to  write  in  the  form  of  a  proportion  the 
two  equal  ratios,  8  :  4  and  6  :  3,  which  both  have  the  same 
value,  2,  we  may  employ  one  of  the  three  following  forms: 

8    :    4  ::  6    :    3  (1) 

8:4=6:3  (2) 


47.  The  first  form  is  the  one  most  extensively  used,  by 
reason  of  its  having  been  exclusively  employed  in  all  the 
older  works  on  mathematics.  The  second  and  third  forms 
are  being  adopted  by  all  modern  writers  on  mathematical 
subjects,  and  in  time  will  probably  entirely  supersede  the 
first  form.  In  this  arithmetic  we  shall  adopt  the  second 
form,  unless  some  statement  can  be  made  clearer  by  using 
the  third  form. 


*  This  evidently  does  not  alter  the  value  of  the  ratio,  since  by  the 
laws  of  fractions,  both  numerator  and  denominator  may  be  divided  by 
the  same  number  without  changing  the  value  of  the  fraction. 


44  ARITHMETIC.  §  2 

48.  A  proportion   may  be  read  in  two  ways.     The  old 
way  to  read  the  above  proportion  was  :  8  is  to  J^as  6  is  to  3  ; 
the  new  way  is :  the  ratio  of  '8  to  4  equals  the  ratio  of  6  to  3. 
The  student  may  read  it  either  way,  but  we  recommend  the 
latter. 

49.  Each  ratio  of  a  proportion  is  termed  a  couplet.     In 
the  above  proportion,  8  :  4  is  a  couplet,  and  so  is  G  :  3. 

50.  The    numbers    forming   the    proportion    are    called 
terms ;  and  they  are  numbered  consecutively  from  left  to 

right,  thus: 

first  second  third  fourth        ; 

8:4=6     :     3 

Hence,  in  any  proportion,  the  ratio  of  the  first  term  to  the 
second  term  equals  the  ratio  of  the  third  term  to  the  fourth 
term. 

51.  The  first  and  fourth  terms  of  a  proportion  are  called 
the  extremes,  and  the  second  and  third  terms  the  means. 
Thus,  in  the  foregoing  proportion,  8  and  3  are  the  extremes 
and  4  and  6  are  the  means. 

52.  A  direct  proportion  is  one  in  which  both  couplets 
are  direct  ratios. 

53.  An  inverse  proportion  is  one  which  requires  one 
of  the  couplets  to  be  expressed  as  an  inverse  ratio.     Thus, 
8  is  to  4  inversely  as  3  is  to  6  must  be  written  8  :  4  =  G  :  3 ; 
i.  e.,  the  second  ratio  (couplet)  must  be  inverted. 

54.  Proportion  forms  one  of  the  most  useful  sections  of 
arithmetic.     In  our  grandfathers'  arithmetics,  it  was  called 
"The  rule  of  three." 

55.  Rule. — In     any    proportion,    the    product    of    the 
extremes  equals  the  product  of  the  means. 

Thus,  in  the  proportion, 

17  :  51  =  14  :  42. 
17  X  42  =  51  X  14,  since  both  products  equal  714. 

56.  Rule. —  The  product    of  the    extremes    divided   by 
either  mean  gives  the  other  mean. 


§  2  ARITHMETIC.  45 

EXAMPLE. — What  is  the  third  term  of  the  proportion  17  :  51  =      :  42  ? 

SOLUTION. — Applying  the  rule,  17  X  42  =  714,  and  714  -*-  51  =  14. 

Ans. 

57.  Rule. —  The  product  of  the  means  divided  by  either 
extreme  gives  the  other  extreme. 

EXAMPLE. — What  is  the  first  term  of  the  proportion     :  51  =  14  :,42  ? 

SOLUTION.— Applying  the  rule,  51  X  14  =  714,  and  714  -=-  42  =  17. 

Ans. 

58.  When  stating  a  proportion  in  which  one  of  the  terms 
is  unknown,  represent  the  missing  term  by  a  letter,  as  x. 
Thus,  the  last  example  would  be  written 

x  :  51  =  14  :  42, 

51  X  14 

and  for  the  value  of  x  we  have  x  = — =  17. 

42 

59.  The  principle   of   all  calculations  in   proportion   is 
this  :   Three  of  the  terms  are  always  given  and  the  remain- 
ing one  is  to  be  found. 

GO.  EXAMPLE. — If  4  men  can  earn  $25  in  one  week,  how  much 
can  12  men  earn  in  the  same  time  ? 

SOLUTION. — The  required  term  must  bear  the  same  relation  to  the 
given  term  of  the  same  kind  as  one  of  the  remaining  terms  bears  to 
the  other  remaining  term.  We  can  then  form  a  proportion  by  which 
the  required  term  may  be  found. 

The  first  question  the  student  must  ask  himself  in  every  calculation 
by  proportion  is:  "What  is  it  I  want  to  find?"  In  this  case  it  is 
dollars.  We  have  two  sets  of  men,  one  set  earning  $25,  and  we  want 
to  know  how  many  dollars  the  other  set  earns.  It  is  evident  that  the 
amount  12  men  earn  bears  the  same  relation  to  the  amount  that  4  men 
earn  as  12  men  bears  to  4  men.  Hence,  we  have  the  proportion,  the 
amount  12  men  earn  is  to  $25  as  12  men  is  to  4  men ;  or,  since  either 
extreme  equals  the  product  of  the  means  divided  by  the  other  extreme, 
we  have 

The  amount  12  men  earn  :  $25  =  12  men  :  4  men,  ,. 

ffi»opj  v  1 2 
or  the  amount  12  men  earn  = j =  $75.     Ans. 

Since  it  matters  not  which  place  x,  or  the  required  term,  occupies, 
the  problem  could  be  stated  in  any  of  the  following  forms,  the  value  of 
x  being  the  same  in  each; 


46 


ARITHMETIC. 


(a)  $25  :  the  amount  12  men  earn  =  4  men  :  12  men;  or  the  amount 

q»OP\  y  -10 

12  men  earn  =  - — ^ — ,  or  $75,  since  either  mean  equals  the  product 

of  the  extremes  divided  by  the  other  mean. 

(b)  4  men  :  12  men  =  $25  :  the  amount  12  men  earn;   or  the  amount 

0>OP)   -y   1  O 

that  12  earn  =  —      — ,  or  $75,  since  either  extreme  equals  the  product 
of  the  means  divided  by  the  other  extreme. 

(c)  12  men  :  4  men  =  the  amount  12  men  earn  :  §25 ;  or  the  amount 
that  12  men  earn  =      '  * — ,  or  $75,  since  either  mean  equals    the 
product  of  the  extremes  divided  by  the  other  mean. 

61.  If  the  proportion  is  an  inverse  one,  first  form  it  as 
though  it  were  a  direct  proportion  and  then  invert  one  of 
the  couplets. 


62. 


EXAMPLES   FOR  PRACTICE. 

Find  the  value  of  x  in  each  of  the  following: 


(a)  $16  :  $64  =  x :  $4.  f  (a)  =  $1. 

(£)  x :  85  =  10  :  17.  (b)  -  50. 

(c)  24  :  x  =  15  :  40.  (c}  =  64. 

(d)  18  :  94  =  2  :  x.                       Ans.  J  (d)  =  lOf. 
(«?)  $75  :  $100  =  x :  100.  (e)  —  75. 
(/)  15  pwt.  :  x  =  21  :  10.  (/)  =  7|  pwt. 
(g~)  x  :  75  yd.  =  §15  :  $5.  [  (g)  =  225  yd. 

1.  If  75  pounds  of  lead  cost  $2.10,  what  would  125  pounds  cost  at  the 
same  rate?  Ans.  $3.50. 

2.  If  A  does  a  piece  of  work  in  4  days  and  B  does  it  in  7  days,  how 
long  will  it  take  A  to  do  what  B  does  in  63  days  1  Ans.  36  days. 

3.  The  circumferences  of  any  two  circles  are  to  each  other  as  their 
diameters.    If  the  circumference  of  a  circle  7  inches  in  diameter  is 
22  inches,  what  will  be  the  circumference  of  a  circle  31   inches  in 
diameter?  Ans.  97f  in. 


INVERSE    PROPORTION. 

63.  In  Art.  53,  an  inverse  proportion  was  defined  as  one 
which  required  one  of  the  couplets  to  be  expressed  as  an 
inverse  ratio.  Sometimes  the  word  inverse  occurs  in  the 
statement  of  the  example;  in  such  cases  the  proportion  can 
be  written  directly,  merely  inverting  one  of  the  couplets. 


§  2  ARITHMETIC.  47 

But  it  frequently  happens  that  only  by  carefully  studying 
the  conditions  of  the  example  can  it  be  ascertained  whether 
the  proportion  is  direct  or  inverse.  When  in  doubt,  the 
student  can  always  satisfy  himself  as  to  whether  the  propor- 
tion is  direct  or  inverse  by  first  ascertaining  what  is  required, 
and  stating  the  proportion  as  a  direct  proportion.  Then,  in 
order  that  the  proportion  may  be  true,  if  the  first  term  is 
smaller  than  the  second  term,  the  third  term  must  be  smaller 
than  the  fourth  ;  or  if  the  first  term  is  larger  than  the  second 
term,  the  tJiird  term  must  be  larger  than  the  fourth  term. 
Keeping  this  in  mind,  the  student  can  always  tell  whether 
the  required  term  will  be  larger  or  smaller  than  the  other  term 
of  the  couplet  to  which  the  required  term  belongs.  Having 
determined  this,  the  student  then  refers  to  the  example  and 
ascertains  from  its  conditions  whether  the  required  term  is 
to  be  larger  or  smaller  than  the  other  term  of  the  same 
kind.  If  the  two  determinations  agree,  the  proportion  is 
direct;  otherwise,  it  is  inverse,  and  one  of  the  couplets  must 
be  inverted. 

64.     EXAMPLE. — A's  rate  of  doing  work  is  to  B's  as  5:7;  if  A 
does  a  piece  of  work  in  42  days,  in  what  time  will  B  do  it  ? 

SOLUTION.— The  required  term' is  the  number  of  days  it  will  take  B 
to  do  the  work.     Hence,  stating  as  a  direct  proportion, 

5  :  7  =  42  :  x. 

Now,  since  7  is  greater  than  5,  x  will  be  greater  than  42.  But,  refer- 
ring to  the  statement  of  the  example,  it  is  easy  to  see  that  B  works 
faster  than  A ;  hence  it  will  take  B  a  less  number  of  days  to  do  the 
work  than  A.  Therefore,  the  proportion  is  an  inverse  one,  and  should 
be  stated 

5  :  7  =  x  :  42, 


from  which  x  =  5  *42  =  30  days.     Ans. 


Had  the  example  been  stated  thus:  The  time  that  A  requires  to  do 
a  piece  of  work  is  to  the  time  that  B  requires,  as  5  :  7 ;  A  can  do  it  in 
42  days,  in  what  time  can  B  do  it  ?  it  is  evident  that  it  would  take  B 
a  longer  time  to  do  the  work  than  it  would  A;  hence,  x  would  be 
greater  than  42,  and  the  proportion  would  be  direct,  the  value  of  x  being 

*4*?  =  68.8  days. 


48  ARITHMETIC.  §  2 

EXAMPLES   FOR   PRACTICE. 

65.  Solve  the  following: 

1.  If  a  pump  which  discharges  4  gal.  of  water  per  min.  can  fill 
a  tank  in  20  hr.,  how  long  will  it  take  a  pump  discharging  12  gal.  per 
min.  to  fill  it  ?  Ans.  6£  hr. 

2.  The  circular  seam  of  a  boiler  requires  50  rivets  when  the  pitch 
is  2-J-  in. ;  how  many  would  be  required  if  the  pitch  were  3£  in.  ? 

Ans.  40. 

3.  The  spring  hangers  on  a  certain  locomotive  are  2-J  in.  wide  and 
f  in.  thick ;  those  on  another  engine  are  of  same  sectional  area,  but 
are  3  in.  wide;  how  thick  are  they  ?  Ans.  f  in. 

4.  A  locomotive  with  driving  wheels  16  ft.  in  circumference  runs  a 
certain  distance  in  5,000  revolutions;  how  many  revolutions  would  it 
make  in  going  the  same  distance  if  the  wheels  were  22  ft.  in  circum- 
ference (no  allowance  for  slip  being  made  in  either  case)  ? 

Ans.  3,636^-  rev. 

UNIT  METHOD. 

66.  In  the  older  books  on  arithmetic,  a  large  number  of 
problems  were  solved  by  proportion ;  but  these  problems  can 
be  solved  much  more  easily  by  the  unit  method,  which  we 
now  proceed  to  explain  by  means  of  examples. 

EXAMPLE  1. — If  a  pump  discharging  4  gallons  of  water  per  minute 
can  fill  a  tank  in  20  hours,  how  long  will  it  take  a  pump  discharging 
12  gallons  per  minute  to  fill  the  tank  ? 

SOLUTION.— A  pump  discharging  4  gallons  per  minute  fills  the  tank 
in  20  hours.  Therefore,  a  pump  discharging  1  gallon  per  minute  fills 
it  in  4  X  20  hours.  Hence,  a  pump  discharging  12  gallons  per  minute 

fills  it  in  4x2?0h°urs  =  20h°UrS  =  6|  hr.     Ans. 
1*  o 

EXAMPLE  2.— If  4  men  earn  $65.80  in  7  days,  how  much  can  14  men, 
paid  at  the  same  rate,  earn  in  12  days  ? 

SOLUTION.—  4  men  in  7  days  earn  $65.80. 

Therefore,  1  man  in  7  days  earns  — ^ — . 

Therefore,  1  man  in  1  day  earns       '     . 

4  X  • 

Therefore,      1  man  in  12  days  earns  $65 .8°  *  12. 

4  X  ' 

Therefore,  14  men  in  12  days  earn  S65-80^  *2  X  14 
Canceling, 
14  men  in  12  days  earn  $65.80  X  3  X  2  =  $65.80  X  6  -  $.394.80.     Ans, 


§  2  ARITHMETIC.  49 

67.  The  student  will  notice  that  in  the  solution  of  these 
examples,  in  the  successive  steps,  the  operations  of  mul- 
tiplication and  division  were  merely  indicated,  and  no 
multiplication  or  division  was  performed  until  the  very  last, 
and  then  the  answer  was  'obtained  easily  by  cancelation. 
In  arithmetical  calculations,  the  student  should  make  it  an 
invariable  habit  to  indicate  the  multiplications  and  divisions 
that  occur  in  the  successive  steps  of  a  solution,  and  not  to 
perform  these  operations  until  the  very  last.  Then,  he  will 
probably  be  able  to  use  the  principle  of  cancelation. 

EXAMPLE.— If  a  block  of  granite  8  feet  long,  5  feet  wide,  and  3  feet 
thick  weighs  7,200  pounds,  what  is  the  weight  of  a  block  of  granite 
12  feet  long,  8  feet  wide,  and  5  feet  thick  ? 

SOLUTION.— If  a  block  8  feet  long,  5  feet  wide,  and  3  feet  thick 
weighs  7,200  pounds,  a  block  1  foot  long,  5  feet  wide,  and  3  feet 

thick  weighs  -^—  pounds ;  a  block  1  foot  long,  1  foot  wide,  and  3  feet 

o 
7  900 

thick  weighs  0^—5;  and  a  block  1  foot  long,  1  foot  wide,  and  1  foot 

7  200 
thick  weighs  5—       — 5  pounds.     Therefore,  by  the  same  reasoning,  a 

"  X  5  X  o 
block  12  feet  long,  8  feet  wide,  and  5  feet  thick  weighs 

7.«0gx  12  X  8  X  «  ^  =  T.«0  X  ff  X  >  X  ,  =  ^  ,b      An, 


EXAMPLES  FOB  PRACTICE. 

1.  If  a  pump  discharges  90,000  gallons  of  water  in  20  hours,  in  what 
time  will  it  discharge  144,000  gallons  ?  Ans.  32  hr. 

2.  When  the  barometer  stands  at  30  inches,  the  pressure  of  the 
atmosphere  is  14.7  pounds  per  square  inch.     What  is  the  atmospheric 
pressure  per  square  inch  when  the  barometer  stands  at  29.5  inches? 
Give  answer  correct  to  three  figures.  Ans.  14.5. 


//.  -Sf.     /.— 8 


MENSURATION  AND  USE  OF 
LETTERS  IN  FORMULAS. 


FORMULAS. 

1.  The  term  formula,  as  used  in  mathematics  and  in 
technical  books,  may  be  defined  as  a  rule  in  which  symbols 
are  used  instead  of  words  ;  in  fact,  a  formula  may  be  regarded 
as  a  shorthand  method  of  expressing  a  rule.     Any  formula 
can  be  expressed  in  words,  and  when  so  expressed  it  becomes 
a  rule. 

Formulas  are  much  more  convenient  than  rules;  they 
show  at  a  glance  all  the  operations  that  are  to  be  performed ; 
they  do  not  require  to  be  read  three  or  four  times,  as  is  the 
case  with  most  rules,  to  enable  one  to  understand  their 
meaning;  they  take  up  much  less  space,  both  in  the  printed 
book  and  in  one's  note  book,  than  rules;  in  short,  whenever 
a  rule  can  be  expressed  as  a  formula,  the  formula  is  to  be 
preferred. 

As  the  term  "  quantity  "  is  a  very  convenient  one  to  use, 
we  will  define  it.  In  mathematics,  the  word  quantity  is 
applied  to  anything  that  it  is  desired  to  subject  to  the  ordi- 
nary operations  of  addition,  subtraction,  multiplication,  etc., 
when  we  do  not  wish  to  be  more  specific  and  state  exactly 
what  the  thing  rs.  Thus,  we  can  say  "two  or  more  num- 
bers," or  "two  or  more  quantities";  the  word  quantity  is 
more  general  in  its  meaning  than  the  word  number. 

2.  The  signs    used  in  formulas   are  the  ordinary  signs 
indicative  of  operations  and  the  signs  of  aggregation.     All 

§3 

For  notice  of  copyright,  see  page  immediately  following  the  title  page. 


2  MENSURATION  AND  §  3 

these  signs  are  explained  in  arithmetic,  but  some  of  them 
will  here  be  explained  in  order  to  refresh  the  student's 
memory. 

3.  The  signs  indicative  of  operations  are  six  in  number, 
viz.:  +,  -,  X,  -*-,    |  ,    V- 

Division  is  indicated  by  the  sign  -i- ,  or  by  placing  a  straight 
line  between  the  two  quantities.  Thus,  25  |  17,  25  /  17, 
and  f4  all  indicate  that  25  is  to  be  divided  by  17.  When 
both  quantities  are  placed  on  the  same  horizontal  line,  the 
straight  line  indicates  that  the  quantity  on  the  left  is  to  be 
divided  by  that  on  the  right.  When  one  quantity  is  below 
the  other,  the  straight  line  between  indicates  that  the  quan- 
tity above  the  line  is  to  be  divided  by  the  one  below  it. 

The  sign  (  \/}  indicates  that  some  root  of  the  quantity  on 
the  right  is  to  be  taken ;  it  is  called  the  radical  sign.  To 
indicate  what  root  is  to  be  taken,  a  small  figure,  called  the 
index,  is  placed  within  the  sign,  this  being  always  omitted 
when  the  square  root  is  to  be  indicated.  Thus,  |/25  indi- 
cates that  the  square  root  of  25  is  to  be  taken;  |/25  indicates 
that  the  cube  root  of  25  is  to  be  taken  ;  etc. 

4.  The  signs  of  aggregation  are  four  in  number;  viz., 

1  (  )>  [  ]>  and  I  }»  respectively  called  the  vinculum,  the 

parenthesis,  the  brackets,  and  the  brace ;  they  are  used 
when  it  is  desired  to  indicate  that  all  the  quantities  included 
by  them  are  to  be  subjected  to  the  same  operation.     Thus, 
if  we  desire  to  indicate  that  the  sum  of  5  and  8  is  to  be  mul- 
tiplied by  7,  and  we  do  not  wish  to  actually  add  5  and  8 
before  indicating  the  multiplication,  we  may  employ   any 
one  of  the  four  signs  of  aggregation  as  here  shown :    5  +  8 
X  7,  (5  +  8)  X  7,  [5  +  8]  X  7,  {5  +  8}  X  7.     The  vinculum  is 
placed  above  those  quantities  which  are  to  be  treated  as  one 
quantity  and  subjected  to  the  same  operations. 

5.  While  any  one  of  the  four  signs  may  be  used  as  shown 
above,  custom  has  restricted  their  use  somewhat.     The  vin- 
culum is  rarely  used  except  in  connection  with  the  radical 
sign.      Thus,   instead   of   writing  y  (5  +  8),  ^  [5  +  8],   or 
y  1 5  -j-  8 1  for  the  cube  root  of  5  plus  8,  all  of  which  would 
be  correct,  the  vinculum  is  nearly  always  used,  4/5  +  8. 


§  3  USE  OF  LETTERS  IN  FORMULAS.  3 

In  cases  where  but  one  sign  of  aggregation  is  needed  (ex- 
cept, of  course,  when  a  root  is  to  be  indicated),  the  paren- 
thesis is  always  used.  Hence,  (5  -f-  8)  X  7  would  be  the 
usual  way  of  expressing  the  product  of  5  plus  8,  and  7. 

If  two  signs  of  aggregation  are  needed,  the  brackets  and 
parenthesis  are  used,  so  as  to  avoid  having  a  parenthesis 
within  a  parenthesis,  the  brackets  being  placed  outside. 
For  example,  [(20  —  5)  -r-  3]  X  9  means  that  the  difference 
between  20  and  5  is  to  be  divided  by  3,  and  this  result  mul- 
tiplied by  9. 

If  three  signs  of  aggregation  are  required,  the  brace, 
brackets,  and  parenthesis  are  used,  the  brace  being  placed 
outside,  the  brackets  next,  and  the  parenthesis  inside.  For 
example,  {  [(20  —  5)  -^  3]  X  9  —  21}  -h  8  means  that  the 
quotient  obtained  by  dividing  the  difference  between  20 
and  5  by  3  is  to  be  multiplied  by  9,  and  that  after  21  has 
been  subtracted  from  the  product  thus  obtained,  the  result 
is  to  be  divided  by  8. 

Should  it  be  necessary  to  use  all  four  of  the  signs  of  aggre- 
gation, the  brace  would  be  put  outside,  the  brackets  next, 
the  parenthesis  next,  and  the  vinculum  inside.  For  example, 
{[(20-5  -T-  3)  X  9  -  21]  -^  8}  X  12. 

6.  As  stated  in  Arithmetic,  when  several  quantities  are 
connected  by  the  various  signs  indicating  addition,  subtrac- 
tion, multiplication,  and  division,  the  operation  indicated  by 
the  sign  of  multiplication  must  always  be  performed  first. 
Thus,  2  +  3x4  equals  14,  3  being  multiplied  by  4  before 
adding  to  2.  Similarly,  10  -4-  2  X  5  equals  1,  since  2x5 
equals  10,  and  10  -f-  10  equals  1.  Hence,  in  the  above  case, 
if  the  brace  were  omitted,  the  result  would  be  £,  whereas, 
by  inserting  the  brace,  the  result  is  36. 

Following  the  sign  of  multiplication  comes  the  sign^of 
division  in  order  of  importance.  For  example,  5  —  9  -f-  3 
equals  2,  9  being  divided  by  3  before  subtracting  from  5. 
The  signs  of  addition  and  subtraction  are  of  equal  value; 
that  is,  if  several  quantities  are  connected  by  plus  and 
minus  signs,  the  indicated  operations  may  be  performed  in 
the  order  in  which  the  quantities  are  placed. 


4  MENSURATION  AND  §  3 

7.  There  is  one  other  sign  used,  which  is  neither  a  sign 
of  aggregation  nor  a  sign  indicative  of  an  operation  to  be 
performed;  it  is  (=),  and  is  called  the  sign  of  equality;  it 
means  that  all  on  one  side  of  it  is  exactly  equal  to  all  on  the 
other  side.    For  example,  2  —  2,  5  —  3  =  2,  5  X  (14  —  9)  =  25. 

8.  Having  called  particular  attention  to  certain  signs 
used  in    formulas,    the   formulas   themselves   will    now   be 
explained.     First,  consider  the  well-known  rule  for  finding 
the  horsepower  of  a  steam  engine,  which  may  be  stated  as 
follows : 

Divide  the  continued  product  of  the  mean  effective  pressure 
in  pounds  per  square  inch,  the  length  of  the  stroke  in  feet,  the 
area  of  tJie  piston  in  square  inches,  and  the  number  of  strokes 
Per  minute,  by  33,000  ;  the  result  will  be  the  horsepower. 

This  is  a  very  simple  rule,  and  very  little,  if  anything, 
will  be  saved  by  expressing  it  as  a  formula,  so  far  as  clear- 
ness is  concerned.  The  formula,  however,  will  occupy  a 
great  deal  less  space,  as  we  shall  show. 

An  examination  of  the  rule  will  show  that  four  quantities 
(viz.,  the  mean  effective  pressure,  the  length  of  the  stroke, 
the  area  of  the  piston,  and  the  number  of  strokes)  are  mul- 
tiplied together,  and  the  result  is  divided  by  33, 000.  Hence, 
the  rule  might  be  expressed  as  follows: 

mean  effective  pressure  stroke 

Horsepower  =   .  .          X  ,.     , 

(in  pounds  per  square  inch)       (in  feet) 

area  of  piston         number  of  strokes 
X  (in  square  inches)  X       (per  minute) 

This  expression  could  be  shortened  by  representing  each 
quantity  by  a  single  letter;  thus,  representing  horsepower 
by  the  letter  "//,"  the  mean  effective  pressure  in  pounds 
per  square  inch  by  "/*,"  the  length  of  stroke  in  feet  by  "  L" 
the  area  of  the  piston  in  square  inches  by  "A,"  the  number 
of  strokes  per  minute -by  "TV,  "and  substituting  these  letters 
for  the  quantities  that  they  represent,  the  above  expression 
would  reduce  to 

Px L X  A  X N 


H  = 


33,000 


§  3  USE  OF  LETTERS  IN  FORMULAS.  5 

a  much  simpler  and  shorter  expression.     This  last  expres- 
sion is  called  a  formula. 

9.  The  formula  just  given 'shows,  as  we  stated  in  the 
beginning,  that   a  formula   is  really  a   shorthand  method  of 
expressing  a  rule.      It  is  customary,  however,  to  omit  the  sign 
of  multiplication  between  two  or  more  quantities  when  they 
are  to  be   multiplied  together,  or  between  a  number  and  a 
letter  representing  a  quantity,  it  being  always  understood 
that,  when  two  letters  are  adjacent  with  no  sign  between 
them,  the  quantities  represented   by  these  letters  are  to  be 
multiplied.      Bearing  this  fact    in  mind,    the  formula    just 
given  can  be  further  simplified  to 

PLAN 
33,000  ' 

10.  The   sign  of    multiplication,    evidently,    cannot   be 
omitted  between  two  or  more  numbers,  as  it  would  then   be 
impossible  to  distinguish  the  numbers.     A  near  approach  to 
this,  however,  may  be  attained  by  placing  a  dot  between  the 
numbers  which   are  to  be   multiplied   together,   and  this  is 
frequently  done  in  works  on  mathematics  when  it  is  desired 
to  economize  space.      In  such  cases  it  is  usual  to  put  the  dot 
higher  than   the  position    occupied   by   the  decimal    point. 
Thus,  2-3  means  the  same  as  2  X  3;  542-749-1,006  indicates 
that  the  numbers  542,  749,  and  1,006  are  to  be  multiplied 
together. 

It  is  also  customary  to  omit  the  sign  of  multiplication  in 
expressions  similar  to  the  following:  a  X  tfb-\-c,  3  X  (b  +  <0, 
(b  +  c)  X  a,  etc.,  writing  them  a  \/b  +  c,  3  (b  +  c),  (b  +  c)  a, 
etc.  The  sign  is  not  omitted  when  several  quantities  are 
included  by  a  vinculum  and  it  is  desired  to  indicate  that 
the  quantities  so  included  are  to  be  multiplied  by  another 
quantity.  For  example,  3  X  ^  +  £,  b  +  c  X  a,  \/b  +  c  y.  a, 
etc.  are  always  written  as  here  printed. 

11.  Before  proceeding  further,  we  will  explain  one  other 
device   that  is   used  by  formula  makers  and  which    is   apt 
to  puzzle  one  who  encounters  it  for  the   first  time — it    is 
the  use  of  what  mathematicians  call  primes  and  subs.,  and 


6  MENSURATION  AND  §  3 

what  printers  call  superior  and  inferior  characters.  As  a 
rule,  formula  makers  designate  quantities  by  the  initial 
letters  of  the  names  of  the'  quantities.  For  example,  they 
represent  volume  by  v,  pressure  by  /,  height  by  //,  etc. 
This  practice  is  to  be  commended,  as  the  letter  itself  serves 
in  many  cases  to  identify  the  quantity  which  it  represents. 
Some  authors  carry  the  practice  a  little  further  and  repre- 
sent all  quantities  of  the  same  nature  by  the  same  letter 
throughout  the  book,  always  having  the  same  letter  repre- 
sent the  same  thing.  Now,  this  practice  necessitates  the 
use  of  the  primes  and  subs,  above  mentioned  when  two 
quantities  have  the  same  name  but  represent  different  things. 
Thus,  consider  the  word  pressure  as  applied  to  steam  at  dif- 
ferent stages  between  the  boiler  and  the  condenser.  First, 
there  is  absolute  pressure,  which  is  equal  to  the  gauge  pres- 
sure in  pounds  per  square  inch  plus  the  pressure  indicated 
by  the  barometer  reading  (usually  assumed  in  practice  to  be 
14.7  pounds  per  square  inch,  when  a  barometer  is  not  at 
hand).  If  this  be  represented  by/,  how  shall  we  represent 
the  gauge  pressure  ?  Since  the  absolute  pressure  is  always 
greater  than  the  gauge  pressure,  suppose  we  decide  to  repre- 
sent it  by  a  capital  letter  and  the  gauge  pressure  by  a  small 
(lower-case)  letter.  Doing  so,  P  represents  absolute  pres- 
sure and  /,  gauge  pressure.  Further,  there  is  usually  a 
"drop  "  in  pressure  between  the  boiler  and  the  engine,  so 
that  the  initial  pressure,  or  pressure  at  the  beginning  of  the 
stroke,  is  less  than  the  pressure  at  the  boiler.  How  shall 
we  represent  the  initial  pressure  ?  We  may  do  this  in  one 
of  three  ways  and  still  retain  the  letter  /or  Pto  represent 
the  word  pressure:  First,  by  the  use  of  the  prime  mark; 
thus,  /'  or  P'  (read  /  prime  and/3  major  prime]  may  be  con- 
sidered to  represent  the  initial  gauge  pressure  or  the  initial 
absolute  pressure.  Second,  by  the  use  of  sub.  figures ;  thus, 
/,  or  Pl  (read/  sub.  one  and  P  major  sub.  one).  Third,  by 
the  use  of  sub.  letters ;  thus,  /,-  or  Pt  (read  /  sub.  i  and  P 
major  sub.  i).  In  the  same  manner  /"  (read  /  second},  /2,  or 
/r  might  be  used  to  represent  the  gauge  pressure  at  release, 
etc.  The  sub.  letters  have  the  advantage  of  still  further 
identifying  the  quantity  represented:  in  many  instances, 


§  3  USE  OF  LETTERS  IN  FORMULAS.  7 

however,  it  is  not  convenient  to  use  them,  in  which  case 
primes  and  subs,  are  used  instead.  The  prime  notation 
may  be  continued  as  follows:  /'",  piv,  pv,  etc.;  it  is  inad- 
visable to  use  superior  figures,  for  example,  /',  /",  p*,  /°, 
etc.,  as  they  are  liable  to  be  mistaken  for  exponents. 

12.  The  main  thing  to  be  remembered  by  the  student  is 
that  when  a  formula  is  given  in  which  the  same  letters  occur 
several  times,  all  like  letters  having  the  same  primes  or  subs. 
represent  the  same  quantities,  while  those  which  differ  in  any 
respect  represent  different  quantities.  Thus,  in  the  formula 


.  t  _. 


w,,  wv  and  w3  represent  the  weights  of  three  different 
bodies;  slt  sy,  and  ss,  their  specific  heats;  and  /15  /2,  and  /s, 
their  temperatures;  while  t  represents  the  final  temperature 
after  the  bodies  have  been  mixed  together.  It  should  be 
noted  that  those  letters  having  the  same  subs,  refer  to  the 
same  bodies.  Thus,  w,,  s^,  and  t^  all  refer  to  one  of  the 
three  bodies;  w2,  s^  t^  to  another  body,  etc. 

It  is  very  easy  to  apply  the  above  formula  when  the 
values  of  the  quantities  represented  by  the  different  letters 
are  known.  All  that  is  required  is  to  substitute  the  numer- 
ical values  of  the  letters  and  then  perform  the  indicated 
operations.  Thus,  suppose  that  the  values  of  «;„  slt  and  ^ 
are,  respectively,  2  pounds,  .0951,  and  80°;  of  «;„  st,  and  ts, 
7.8  pounds,  1,  and  80°;  and  of  zu3,  s3,  and  A,,  3£  pounds, 
.1138,  and  780°;  then,  the  final  temperature  t  is,  substi- 
tuting these  values  for  their  respective  letters  in  -the 
formula, 

2  X  .0951  X  80  +  7.8  X  1  X  80  +  3j  X  .1138  X  780 

2  X  .0951  +  7.8  X  1  -f  3±  X  .1138 
_  15.216  +  624  +  288.483  _  927.  699 
.  1902  +  7.  8  +  .  36985    ~  8.  36005 

In  substituting  the  numerical  values,  the  signs  of  multi- 
plication are,  of  course,  written  in  their  proper  places;  all 
the  multiplications  are  performed  before  adding,  according 
to  the  rule  previously  given. 


8  MENSURATION  AND  §  3 

13.  The  student  should  now  be  able  to  apply  any  for- 
mula involving  only  algebraic  expressions  that  he  may  meet 
with,  and  which  does  not  require  the  use  of  logarithms  for 
its  solution.  We  will,  however,  call  his  attention  to  one  or 
two  other  facts  that  he  may  have  forgotten. 

Expressions  similar  to  -7-7-  sometimes  occur,  the  heavy  line 

25 

indicating  that  160  is  to  be  divided  by  the  quotient  obtained 
by  dividing  660  by  25.  If  both  lines  were  light,  it  would 
be  impossible  to  tell  whether  160  was  to  be  divided  by 

—  —  ,  or  whether  -  was  to  be  divided  by  25.      If  this  latter 

160 

result  were  desired,  the  expression  would  be  written——.     In 

every  case  the  heavy  line  indicates  that  all  above  it  is  to  be 
divided  by  all  below  it. 

In   an  expression   like   the   following,  —       —  -,  the  heavy 


line    is   not   necessary,    since    it    is   impossible    to    mistake 
the  operation  that  is  required  to  be  performed.     But,  since 

,   660       175  +  600    .,  175  -f  660  ,  ,   660 

7  +  -25-=     -35—  ,  ^  we  substitute—^—  for  7  +  —, 

the  heavy  line   becomes  necessary   in  order    to   make   the 
resulting  expression  clear.     Thus, 

160  160        _  160 

660  ~~  175  +  660    •  835" 
+  25  25  25 

14.  Fractional  exponents  are  sometimes  used  instead  of 
the  radical  sign.  That  is,  instead  of  indicating  the  square, 
cube,  fourth  root,  etc.  of  some  quantity,  as  37,  by  4/37, 
1/37^4/37^  etc.,  these  roots  are  indicated  by  37',  37§,  371, 
etc.  Should  the  numerator  of  the  fractional  exponent  be 
some  quantity  other  than  1,  this  quantity,  whatever  it  may 
be,  indicates  that  the  quantity  affected  by  the  exponent  is 
to  be  raised  to  the  power  indicated  by  the  numerator;  the 


§  3  USE  OF  LETTERS  IN  FORMULAS.  9 

denominator  is  always  the  index  of  the  root.  Hence,  instead 
of  writing  |/37"  for  the  cube  root  of  the  square  of  37,  it  may 
be  written  37?,  the  denominator  being  the  index  of  the  root  ; 
in  other  words,  |/372  =  37s.  Likewise,  |/(1  -(-  a*  b}%  may  also 
be  written  (1  -f-  a'1  /£)*,  a  much  simpler  expression. 

15.  We  will  now  give  several  examples  showing  how  to 
apply  some  of  the  more  difficult  formulas  that  the  student 
may  encounter. 

1.  The  area  of  any  segment.  of  a  circle  that  is  less  than 
(or  equal  to)  a  semicircle  is  expressed  by  the  formula 

TT  r1  E      c  ,         ,  . 

^  =  "sir  -s  <'-*>• 

in  which  A  =  area  of  segment; 
TT  =  3.1416; 
r  =  radius; 
E  =  angle  obtained  by  drawing  lines  from  the  cen- 

ter to  the  extremities  of  arc  of  segment  ; 
c  =  chord  of  segment  ; 
1i  =  height  of  segment. 

•  EXAMPLE.  —  What  is  the  area  of  a  segment  whose  chord  is  10  inches 
long,  angle  subtended  by  chord  is  83.46°,  radius  is  7.5  inches,  and 
height  of  segment  is  1.91  inches? 

SOLUTION.  —  Applying  the  formula  just  given, 


=  40.968  -  27.95  .-  13.018  sq.  in.,  nearly.     Ans. 

2.  The  area  of  any  triangle  may  be  found  by  means  of 
the  following  formula,  in  which  A  =  the  area,  and  a,  b,  and  c 
represent  the  lengths  of  the  sides: 


EXAMPLE. — What  is  the  area  of  a  triangle  whose  sides  are  21  feet, 
46  feet,  and  50  feet  long  ? 

SOLUTION. — In  order  to  apply  the  formula,  suppose  we  let  a  repre- 
sent the  side  that  is  21  feet  long;  b,  the  side  that  is  50  feet  long;  andr, 
the  side  that  is  46  feet  long.  Then,  substituting  in  the  formula 


10  MENSURATION  AND 


=  25  |/441  -  8.25*  =  25  4/441  -  68.0625  =  25  |/372.9375 

=  25  X  19.312  =  482.8  sq.  ft.,  nearly.     Ans. 

The  operations  in  the  above  examples  have  been  extended 
much  farther  than  was  necessary ;  it  was  done  in  order  to 
show  the  student  every  step  of  the  process.  The  last  for- 
mula is  perfectly  general,  and  the  same  answer  would  have 
been  obtained  had  the  50-foot  side  been  represented  by  a, 
the  46-foot  side  by  £,  and  the  21-foot  side  by  c. 

3..  The  Rankine-Gordon  formula  for  determining  the 
least  load  in  pounds  that  will  cause  a  long  column  to  break  is 

D_     SA 

r> 


in  which  P  —  load  (pressure)  in  pounds  ; 

5  —  ultimate  strength  (in  pounds  per  square  inch) 

of  the  material  composing  the  column  ; 
A  =  area    of    cross-section    of     column    in    square 

inches  ; 

q  =  a  factor  (multiplier)  whose  value  depends  upon 
the  shape  of  the  ends  of  the  column  and  on  the 
material  composing  the  column  ; 
/  =  length  of  column  in  inches; 

G  ==  least    radius    of    gyration    of   cross-section   of 
column. 

The  values  of  5,  g,  and  G*  are  given  in  printed  tables  in 
books  in  which  this  formula  occurs. 

EXAMPLE.  —  What  is  the  least  load  that  will  break  a  hollow  steel 
column  whose  outside  diameter  is  14  inches;  inside  diameter,  11  inches; 
length,  20  feet,  and  whose  ends  are  flat  ? 

SOLUTION.—  For  steel,    5  =  150,000,    and  q  =  --.   for  flat-ended 


steel  columns;    A,  the  area  of  the  cross-section,  =  .7854  (</,*  —  dj) 
=  .7854(14*  —  II2),  </!  and  d*  being  the  outside  and  inside  diameters, 


§3  USE  .OF  LETTERS  IN  FORMULAS.  11 

respectively ;   /  =  20  X  12  =  240  inches ;  and  G*  —    '    ~^~     Q  —  jj.      ". 

Substituting  these  values  in  the  formula 

_      SA      _  150,000  X  .7854  (14*  —  11s) 
~"  1          ZI  ~  ~i  1  240* 

+  q  G~*  +  25,000  X  14*  +  11* 

16 

150.000  X  58.905      8.835.750      7Q1-9111.        . 
1  +  .1163          =    1.1163     =  7'915'211 

4.      EXAJVIPLE. — When  A  =  10,   B  =  8,  C  =  5,  and  D  =  4,  what  is 
the  value  of  E  in  the  following  ? 


3  /      BCD 
(a)    E=A/  — -p-; 

y  -« (»+£•) 

SOLUTION. — (a)  Substituting, 


X5X4 

Fl)' 

To  simplify  the  denominator,  square  the  4  and  5,  add  the  resulting 
fraction  to  2,  and  multiply  by  10.     Simplifying,  we  have 


" ^ 7T7r~  —  ^ '*"  —  \  /  fifiO  —  r    "R^"' 

r      ~<>K 


'(2  +  25)        T      ~~25        *       25 
Reducing  the  fraction  to  a  decimal  before  extracting  the  cube  root, 

E  =  ^6.0606  =  1.823.     Ans. 
(&)    Substituting, 


10-1/4 

16.  In  the  preceding  pages,  the  unknown  quantity  has 
always  been  represented  by  the  single  letter  at  the  left 
of  the  sign  of  equality,  while  the  letters  at  the  right  have 
represented  known  values  from  which  the  required  values 
could  be  found.  It  is  possible,  however,  to  find  the  value  of 
the  quantity  represented  by  any  letter  in  a  formula,  if  the 
values  represented  by  all  the  others  are  known.  For  example, 
let  it  be  required  to  find  how  many  strokes  per  minute  an 


12  MENSURATION  AND  §  3 

engine  having  a  piston  area  of  78.54  square  inches  must 
make  in  order  to  develop  60  horsepower,  if  the  mean  effective 
pressure  is  40  pounds  per  square  inch  and  the  length  of 
stroke  is  1£  ft.  By  substituting  the  given  values  in  the  for- 
.  PLAN 


40  X  Ij-X  78.54  X  N 

33,000 
in  which  N,  the  number  of  strokes,  is  to  be  found. 

But  it  is  evident  that  the  expression  on  the  right  of  the 

sign  of  equality  is  equal  to  -  —  ^  -  '•  —  X  N,  a  fraction 

oo,000 

whose  numerator  is  composed  of  three  factors.  Reducing 
the  numerator  to  a  single  number  by  performing  the  indi- 
cated multiplications,  we  obtain,  after  canceling, 


If  60  equals  .119  N,  then  TV  equals  60  divided  by  .119;  hence, 

N  =  —  —  =  504.2  strokes  per  minute. 
.  119 

The  method  of  procedure  is  essentially  the  same  when  the 
unknown  quantity  occurs  in  the  denominator  of  a  formula. 

111  £72 

Thus,  in  the  formula/=  -  ,  suppose  that/=  375,  /;/  =  1.25, 
and  v  =  60.     Then,  substituting, 

1.25  X  60'       4,500 
o75  =  -  =  -  . 
r  r 

But,  if  375  equals  4,500  divided  by  r,  then  375  X  r  =  4,500; 

hence,  r  must  equal  4,500  divided  by  375,  or  r  =  -^  —  =  12. 

o75 


EXAMPLES  FOR  PRACTICE. 

Find  the  numerical  values  of  x  in  the  following  formulas,  when 
A  —  9,  B  =  8,  d  =  10,  e  =  3,  and  c  =  2  : 


Ans.  x  =  1  J. 


§  3 


USE  OF  LETTERS  IN  FORMULAS. 


13 


Ans.  x  =  29. 

Ans.*  =  8. 

Ans.  jr=12&. 

Ans.  JT=  .396+. 


MENSURATION. 

17.     Mensuration   treats  of   the  measurement  of   lines, 
angles,  surfaces,  and  solids. 


LINES    AND    ANGLES. 

18.  A  straight   line   is   one  that  does  not  change  its 
direction  throughout  its  whole  length.     To  distinguish  one 
straight  line  from  another,  two  of  its 

points  are  designated  by  letters.     The  j B 

line  shown  in  Fig.  1  would  be  called  the  FIG.  i. 

line  A  B. 

19.  A  curved  line  changes  its  di- 
rection at    every  point.     Curved    lines 
are  designated  by  three  or  more  letters, 
as  the  curved  line  ABC,  Fig.  2. 

20.  Parallel    lines    (Fig.     3)    are 
those  which  are  equally  distant   from 
each  other  at  all  points. 

21.  A    line,   is    perpendicular   to 
another  (see  Fig.  4)  when  it  meets  that 
line  so  as  not  to  incline  towards  it  on 
either  side. 

22.  A   vertical   line    is    one    that 

points  towards  the  center  of  the  earth,  j 

and  is  also  known  as  a  plumb-line.  B 

23.  A  horizontal  line  (see  Fig.  5) 

is  one  that  makes  a  right  angle  with  I 

-   ..  Horizontal. 

any  vertical  line.  FIG.  5. 


FIG.  3. 


FIG.  4. 


14 


MENSURATION  AND 


FIG.  6. 


24.  An  angle  is  the  opening  between  two  lines  which 
intersect  or  meet ;  the  point  of  meeting  is  called  the  vertex 

of  the  angle.  Angles  are  distinguished 
by  naming  the  vertex  and  a  point  on  each 
line.  Thus,  in  Fig.  6,  the  angle  formed 
by  the  lines  A  B  and  C  B  is  called  the 
angle  ABC,  or  the  angle  C  B  A ;  the 
letter  at  the  vertex  is  always  placed  in  the  middle.  When 
an  angle  stands  alone  so  that  it  cannot  be  mistaken  for  any 
other  angle,  only  the  vertex  letter  need  be  used.  Thus,  the 
angle  referred  to  might  be  designated  simply  as  the  angle  B. 

25.  If  one  straight  line  meets  an-     J 
other  straight  line  at  a  point  between  its 
ends,  as  in  Fig.   7,  two  angles,  ABC 
and   A  B  D,   are   formed,    which    are    c 

called  adjacent  angles.  FIG. 


26.  When  these  adjacent  angles, 
ABC  and  A  B  D,  are  equal,  as  in 
Fig.  8,  they  are  called  right  angles. 


B 

FIG.  8. 


27.  An  acute  angle  is  less  than  a 
right  angle.  A  £  C,  Fig.  9,  is  an  acute 
angle. 


FIG.  9. 


28.  An  obtuse  angle  is  greater 
than  a  right  angle.  A  B  D  (Fig.  10)  is 
an  obtuse  angle. 


29.  A  circle  (see  Fig.  11)  is  a  figure 
bounded  by  a  curved  line,  called  the  circum- 
ference, every  point  of  which  is  equally  dis- 
tant from  a  point  within,  called  the  center. 


§3 


USE  OF  LETTERS  IN  FORMULAS. 


15 


FIG.  12. 


30.  An  arc  of  a  circle  is  any  part  of  its  circumference ; 
thus  a  e  b,  Fig.  12,  is  an  arc  of  the  circle. 

31.  The  circumference  of  every  circle  is 
considered  to  be  divided  into  360  equal  parts, 
or   arcs,    called   degrees;    every    degree   is 
subdivided  into  GO  equal  parts,  called  min- 
utes, and  every  minute  is  again  divided  into 
60  equal  parts,  called  seconds. 

Since  1  degree  is  7^  of  any  circumference,  it  follows  that 
the  length  of  a  degree  will  be  different  in  circles  of  different 
sizes,  but  the  proportion  of  the  length  of  an  arc  of  1  degree 
to  the  whole  circumference  will  always  be  the  same,  viz., 
^J-g-  of  the  circumference. 

Degrees,  minutes,  and  seconds  are  denoted  by  the  symbols 
°,  ',  '.  Thus,  the  expression  37°  14'  44*  is  read  37  degrees 
14  minutes  44  seconds. 


32.  The  arcs  of  circles  are  used  to  measure  angles.  An 
angle  having  its  vertex  at  the  center 
of  a  circle  is  measured  by  the  arc  in- 
cluded between  its  sides  ;  thus,  in 
Fig.  13,  the  arc  F B  measures  the  angle 
FOB.  If  the  arc  FB  contains  20°, 
or  ^/v  of  the  circumference,  the  angle 
FOB  would  be  an  angle  of  20°;  if  it 
contained  20°  14'  18",  it  would  be  an 
FIG.  13.  angle  of  20°  14'  18",  etc. 

In  the  figure,  if  the  line  C  D  be  drawn  perpendicular  to 
A  B,  the  adjacent  angles  will  be  equal,  and  the  circle  will 
be  divided  into  four  equal  angles,  each  of  which  will  be  a 

right  angle.     A  right  angle,  therefore,  is  an  angle  of  — — , 

or  90° ;  two  right  angles  are  measured  by  180°,  or  half  the 
circumference,  and  four  right  angles  by  the  whole  circum- 
ference, or  360°.  One-half  of  a  right  angle,  as  E  O  B,  is  an 
angle  of  45°.  An  acute  angle  may  now  be  defined  as  an 
angle  of  less  than  90°,  and  an  obtuse  angle  as  one  of  more 


16 


MENSURATION  AND 


than    90°.      These    values    are    important    and    should    be 
remembered. 

33.  From  the  foregoing  it  will  be  evident  that  if  a 
number  of  straight  lines  on  the  same 
side  of  a  given  straight  line  meet  at  the 
same  point,  the  sum  of  all  the  angles 
formed  is  equal  to  two  right  angles,  or 
180°.  Thus,  in  Fig.  14,  angles  COB' 
+DOC+EOD+FOE+AOF 


o 

FIG.  14. 


=  2  right  angles,  or  180°. 

34.  Also,  if  through  a  given  point 
any  number  of  straight  lines  be  drawn, 
the  sum  of  all  the  angles  formed  about 
the  points  of  intersection  equals  four 
right  angles,  or  360°.  Thus,  in  Fig.  15, 
angles/fOF+FOC+COA  +  AOG 
+GOE+EOD+DOB+BOH 
=  4  right  angles,  or  360°. 

EXAMPLE. — In  a  flywheel  with  12  arms,  how  many  degrees  are 
there  in  the  angle  included  between  the  center  lines  of  any  two  arms, 
the  arms  being  spaced  equally  ? 

SOLUTION.— Since  there  are  12  arms,  there  are   12  angles,  which 


together  equal  360°. 


Hence,  one  angle  equals  ^  of  360°,  or  ^-r-  =  30°. 

Ans. 


EXAMPLES  FOR  PRACTICE. 

1.  How  many  seconds  are  in  32°  14'  6"  ?  Ans.  116,046  sec. 

2.  How  many  degrees,  minutes,  and  seconds  do  38,582  seconds 
amount  to?  Ans.  10°  43'  2". 

3.  How  many  right  angles  are  there  in  an  angle  of  170°  ? 

Ans.  If  right  angles. 

4.  In  a  pulley  with  five  arms,  what  part  of  a  right  angle  is  included 
between  the  center  lines  of  any  two  arms  ?         Ans.  |  of  a  right  angle. 

5.  If  one  straight  line  meets  another  so  as  to  form  an  angle  of 
20r'  10',  what  does  its  adjacent  angle  equal  ?  Ans.  159°  50'. 

6.  If  a  number  of  straight  lines  meet  a  given  straight  line  at  a 
given  point,  all  being  on  the  same  side  of  the  given  line,  so  as  to  form 
six  equal  angles,  how  many  degrees  are  there  in  each  angle  ?    Ans.  30°. 


§3 


USE  OF  LETTERS  IN  FORMULAS. 


1? 


QUADRILATERALS. 

35.  A  plane  figure  is  any  part  of  a  plane  or  flat  sur- 
face bounded  by  straight  or  curved  lines. 

36.  A  quadrilateral  is  a  plane  figure  bounded  by  four 
straight  lines. 

37.  A  parallelogram  is  a  quadrilateral  whose  opposite 
sides  are  parallel. 

There  are  four  kinds  of  parallelograms :   the  square,  the 
rectangle,  the  rhombus,  and  the  rhomboid. 

38.  A  rectangle  (Fig.  16)  is  a  parallelo- 
gram whose  angles  are  all  right  angles. 

FIG.  16. 


39.     A    square   (Fig.  17)   is    a    rectangle 
whose  sides  are  all  of  the  same  length. 


FIG.  17. 

4O.  A  rhomboid  (Fig.  18)  is 
a  parallelogram  whose  opposite 
sides  are  equal  and  parallel,  and 
whose  angles  are  not  right  angles. 


41.  A  rhombus  (Fig.  19)  is 
a  parallelogram  having  equal 
sides,  and  whose  angles  are  not 
right  angles. 


FIG.  19. 

42.  A  trapezoid  (Fig.  20)   is 
a  quadrilateral  which  has  only  two 

of  its  sides  parallel.  FIG.  20. 

43.  The  altitude  of  a  parallelogram  or  a  trapezoid  is 
the   perpendicular    distance  between  the  parallel   lines,   as 
shown  by  the  vertical  lines  in  Figs.  18,  19,  and  20. 

44.  The  base  of  any  plane  figure  is  the  side  on  which 
it  is  supposed  to  stand. 


18  MENSURATION  AND  §  3 

45.  The  area  of  a  surface  is  expressed  by  the  number 
of  unit  squares  it  will  contain. 

46.  A  unit  square   is  the  square  having  a  unit  for  its 
side.     For  example,  if  the  unit  is  1  inch,  the  unit  square  is 
the  square  each  of   whose  sides  measures  1   inch  in  length, 
and  the  area  of  a  surface  would  be  expressed  by  the  number 
of  square  inches  it  would  contain.     If  the  unit  were  1  foot, 
the  unit  square  would  measure  1  foot  on  each  side,  and  the 
area  of  the  given  surface  would  be  the  number  of  square 
feet  it  would  contain,  etc. 

The  square  that  measures  1  inch  on  a  side  is  called  a 
square  inch,  and  the  one  that  measures  1  foot  on  a  side 
is  called  a  square  foot.  Square  inch  and  square  foot  are 
abbreviated  to  sq.  in.  and  sq.  ft. 

47.  To  find  the  area  of  any  parallelogram: 
Rule  1. — Multiply  the  base  by  the  altitude. 

NOTE. — Before  multiplying,  the  base  and  altitude  must  be  reduced 
to  the  same  kind  of  units ;  that  is,  if  the  base  should  be  given  in  feet 
and  the  altitude  in  inches,  they  could  not  be  multiplied  together  until 
either  the  altitude  had  been  reduced  to  feet  or  the  base  to  inches. 
This  principle  holds  throughout  the  subject  of  mensuration. 

EXAMPLE  1.— The  sides  of  a  square  piece  of  sheet  iron  are  each 
10^  inches  long.  How  many  square  inches  does  it  contain  ? 

SOLUTION. —  10£  inches  —  10.25  inches  when  reduced  to  a  decimal. 
The  base  and  altitude  are  each  10.25  inches.  Multiplying  them 
together,  10.25  X  10.25  =  105.06+  sq.  in.  Ans. 

EXAMPLE  2.— What  is  the  area  in  square  rods  of  a  piece  of  land  in  the 
shape  of  a  rhomboid,  one  side  of  which  is  8  rods  long  and  whose 
length,  measured  on  a  line  perpendicular  to  this  side,  is  200  feet  ? 

SOLUTION.— The  base  is  8  rods  and  the  altitude  200  feet.  As  the 
answer  is  to  be  in  rods,  the  200  feet  should  be  reduced  to  rods. 

Reducing  200  H-  16$  =  200  H-  y  =  12.12  rods.     Hence,  area  =  8  X  12.12 
=  96.96sq.  rd.     Ans. 

48.  To  find  the  area  of  a  trapezoid: 

Rule  2. — Multiply  one-Jialf  the  sum  of  the  parallel  sides 
by  the  altitude. 

EXAMPLE.— A  board  14  feet  long  is  20  inches  wide  at  one  end  and 
16  inches  wide  at  the  other.  If  the  ends  are  parallel,  how  many  square 
feet  does  the  board  contain  ? 


§  3  USE  OF  LETTERS  IN  FORMULAS.  19 

SOLUTION.— One-half    the    sum    of    the    parallel    sides  =  ^— 

=  18  inches  =  l£  feet.     The  length  of  the  board  corresponds  to  the 
altitude  of  a  trapezoid.     Hence,  14  X  H  =  21  sq.  ft.     Ans. 

49.  Having  given  the  area  of  a  parallelogram  and  one 
dimension,  to  find  the  other  dimension: 

Rule  3. — Divide  the  area  by  the  given  dimension. 

EXAMPLE. — What  is  the  width  of  a  parallelogram  whose  area  is 
212  square  feet  and  whose  length  is  26^  feet  ? 

SOLUTION.—    212  -=-  26^  =  212  -s-  ^-  =  8  ft.     Ans. 

The  following  examples  illustrate  a  few  special  cases: 
EXAMPLE  1. — An  engine  room  is  22  feet  by  32  feet.     The  engine  bed 
occupies  a  space  of  3  feet  by  12  feet;  the  flywheel  pit,  a  space  of  2  feet 
by  6  feet,  and  the  outer  bearing  a  space  of  2  feet  by  4  feet.    How  many 
square  feet  of  flooring  will  be  required  for  the  room  ? 
SOLUTION.— Area  of  engine  bed        =  3  X  12  =  36  sq.  ft. 
Area  of  flywheel  pit      =  2  X    6  =  12  sq.  ft. 
Area  of  outer  bearing  =  2  X    4  =    8  sq.  ft. 

Total,     56  sq.  ft. 

Area  of  engine  room  =  22  X  32  =  704  sq.  ft. 
704  —  56  =  648  sq.  ft.  of  flooring  required.     Ans. 

EXAMPLE  2. — How  many  square  yards  of  plaster  will  it  take  to  cover 
the  sides  and  ceiling  of  a  room  16  X  20  feet  and  11  feet  high,  having 
four  windows,  each  7x4  feet,  and  three  doors,  each  9x4  feet  over  all, 
the  baseboard  coming  6  inches  above  the  floor  ? 
SOLUTION.— 

Area  of  ceiling  =  16  X  20  =  320  sq.  ft. 
Area  of  end  walls  =  2(16  X  11)  =  352  sq.  ft. 
Area  of  side  walls  =  2(20  X  11)  =  440  sq.  ft. 

Total  area  =  1,112  sq.  ft. 
From  the  above  must  be  deducted: 

Windows  =  4(7  X  4)  =  112  sq.  ft. 
Doors        =  3(9  X  4)  =  108  sq.  ft. 

Baseboard  less  the  width  of  three  doors  =  (72  —  12)  X  ^  =  30  sq.  ft. 

Total  number  of  feet  to  be  deducted  =  112  +  108  +  30  =  250  sq.  ft. 
Hence,    number    of    square     feet    to    be    plastered  =  1,112  —  250 
=  862  sq.  ft.,  or  95J  sq.  yd.     Ans. 

EXAMPLE  3. — How  many  acres  are  contained  in  a  rectangular  tract  of 
land  800  rods  long  and  520  rods  wide  ? 

SOLUTION.—  800  X  520  =  416,000  sq.  rd.  Since  there  are  160  square 
rods  in  1  acre,  the  number  of  acres  =  416,000  H-  160  =  2,600  acres.  Ans. 


MENSURATION  AND 


EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  area   in  square  feet  of  a  rhombus  whose  base  is 
84  inches  and  whose  altitude  is  3  feet  ?  Ans.  21  sq.  ft. 

2.  A  flat  roof,  46  feet  by  80  feet  in  size,  is  covered  by  tin  roofing 
weighing  one-half  pound  per  square  foot ;  what  is  the  total  weight  of 
the  roofing  ?  Ans.  1,840  Ib. 

3.  One  side  of  a  room  measures  16  feet.      If  the  floor  contains 
240  square  feet,  what  is  the  length  of  the  other  side  ?  Ans.  15  ft. 

4.  How  many  square  feet  in  a  board  12  feet  long,  18  inches  wide 
at  one  end,  and  12  inches  wide  at  the  other  end  ?  Ans.  15  sq.  ft. 

5.  How  much  would  it  cost  to  lay  a  sidewalk  a  mile  long  and  8  feet 
6  inches  wide,  at  the  rate  of  20  cents  per  square  foot  ?    How  much  at 
the  rate  of  $1.80  per  square  yard  ?  Ans.  $8,976  in  each  case. 

6.  How  many  square  yards  of  plastering  will  be  required  for  the 
ceiling  and  walls  of  a  room  10  ft.  X  15  ft.  and  9  feet  high ;  the  room  con- 
tains one  door  3£  ft.  X  7  ft.,  three  windows  3£  ft.  X  6  ft.,  and  a  baseboard 
8  inches  high  ?  Ans.  53.5  sq.  yd. 

THE    TRIANGLE. 

5O.     A  triangle  is  a  plane  figure  having  three  sides. 

51.  An  isosceles    triangle  is 
one  having  two  of  its  sides  equal ; 
see  Fig.  21. 

52.  An  equilateral   triangle 
(Fig.  22)  is  one  having  all  of  its 

sides  of  the  same  length.  FlG-  33. 

53.     A  scalene  triangle  (Fig.  23)  is  one 
having  no  two  of  its  sides  equal. 
FIG.  23. 

54.  A  right-angled  triangle  (Fig.  24) 
is  any  triangle  having  one  right  angle. 
The  side  opposite  the  right  angle  is  called 
the  hypotenuse.  A  right-angled  triangle 
is  now  usually  called  a  right  triangle. 


55. 


In  any  triangle  the  sum  of  the  three  angles  equals 
two  right  angles,  or  180°.  Thus,  in 
Fig.  25,  the  sum  of  the  angles  A,  B, 
and  ^equals  two  right  angles, or  180°. 
Hence,  if  any  two  angles  of  a  tri- 
•C  angle  are  given  and  it  is  required  to 
find  the  third  angle: 


FIG.  25. 


§  3  USE  OF  LETTERS  IN  FORMULAS.  21 

Rule  4. — Add  the  two  given  angles  and  subtract  their 
sum  from  180° ;  the  remainder  will  be  the  third  angle. 

EXAMPLE. — If  two  angles  of  a  triangle  are  48°  16'  and  47°  50',  what 
does  the  third  angle  equal  ? 

SOLUTION. — First  reduce  48°  16'  and  47°  50'  to  minutes,  for  conve- 
nience in  adding  and  subtracting  the  angles.  48°  =  48  X  60'=  2,880'; 
2,880'  +  16'  =  2,896' ;  hence,  48°  16'  =  2,896'.  In  like  manner,  47°  50' 
=  47  X  CO'  +  50'  =  2,820'+  50'  =  2,870'.  Adding  the  two  angles  and  sub- 
tracting the  sum  from  180°  reduced  to  minutes,  2,896'  +  2,870'  =  5,766'; 
180°  =  180  X  60'=  10,800' ;  10,800  -  5,766  =  5,034'.  Reducing  this  last 

number  to  degrees  and   minutes,     '        =  83f£°  =  83°  54'.    Hence,  the 
third  angle  in  the  triangle  =  83°  54'.     Ans. 

56.  In  any  right  triangle  there  can  be  but  one  right 
angle,  and  since  the  sum  of  all  the  tt 

angles  is  two  right  angles,  it  is  evident 
that  the  sum  of  the  two  acute  angles 
must  equal  one  right  angle,  or  90°. 
Therefore,  if  in  any  right  triangle  one 
acute  angle  is  known,  to  find  the  other 
acute  angle:  FIG.  26. 

Rule  5. — Subtract  the  known  acute  angle  from  90°;  the 
result  will  be  the  other  acute  angle. 

EXAMPLE. — If  one  acute  angle,  as  A,  of  the  right  triangle  ABC, 
Fig.  26,  equals  30°,  what  does  the  angle  B  equal? 
SOLUTION.—    90°  —  30°  =  60'.     Ans. 

57.  If  a  straight  line  be  drawn 
through  two  sides  of  a  triangle,  parallel 
to  the  third  side,  a  second  triangle  will 
be  formed  whose  sides  will  be  propor- 
tional to  the  corresponding  sides  of  the 
first  triangle.  Thus,  in  the  triangle 
ABC,  Fig.  27,  if  the  line  D  E  be  drawn 
parallel  to  the  side  B  C,  the  triangle 
FIG.  27.  v  A  D  E  will  be  formed  and  we  shall  have 

(1)  vSide  A  D  :  side  D  E  =  side  A  B  :  side  B  C ;  and, 

(2)  Side  A  E  :  side  D  E  =  side  A  C  :  side  B  C  ;  also, 

(3)  Side  A  D  :  side  A  E  =  side  A  B  :  side  A  C. 


22 


MENSURATION  AND 


§  3 


FIG- 


EXAMPLE  —In  Fig.  27,  if  A  B  =  24,  B  C  =  18,  and  D  E  =  8,  what 
does  A  D  equal  ? 

SOLUTION.  —  Writing  these  values  for  the  sides  in  (1), 

24-  v  8 

A  D  :  8  =  24  :  18  ;  whence,  A  D  =       *     =  lOf  .     Ans. 

io 

58.     In  any  right  triangle, 
the   square   described    on    the 
hypotenuse  is  equal  to  the  sum 
H  of  the  squares  described  upon 

the  other  two  sides-  If  A  B  C> 
Fig.  28,  is  a  right  triangle 

right-angled  at  .5,  then  the 
square  described  upon  the 
hypotenuse  A  C  is  equal  to 

the  sum  °f  the  squares  de- 

scribed    upon    the    sides   A  B 
an(j      ^  ^7       Hence,     having 
given  the  two  sides  forming 
the  right  angle  in  a  right  triangle,  to  find  the  hypotenuse  : 

Rule  6.  —  Square  each  of  the  sides  forming  the  right  angle  ; 
add  the  squares  together  and  take  the  square  root  of  the  sum. 

EXAMPLE.  —  If  A  B  —  3  inches  and  D  C  —  4  inches,  what  is  the  length 
of  the  hypotenuse  A  C  1 

SOLUTION.—  Squaring  each  of  the  given  sides,  32  =  9  and  42  =  16. 
Taking  the  square  root  of  the  sum  of  9  and  16,  the  hypotenuse 

=  1/9  +  16  =  V'25  =  5  in.     Ans. 

59.  If  the  hypotenuse  and  one  side  are  given,  the  other 
side  can  be  found  as  follows: 

Rule  7.  —  Subtract  the  square  of  the  given  side  from  the 
square  of  the  hypotemise,  and  extract  the  square  root  of  the 
remainder. 

EXAMPLE  1.  —  The  side  given  is  3  inches,  the  hypotenuse  is  5  inches; 
what  is  the  length  of  the  other  side  ? 

SOLUTION.—    32  =  0  ;  5*  =  25.     25  -  9  =  16,  and  4/16  =  4  in.     Ans. 


§3 


USE  OF  LETTERS  IN  FORMULAS. 


23 


150 


EXAMPLE  2. — If  from  a  church  steeple  which  is  150  feet  high  a  rope 
is  to  be  attached  to  the  top  and  to  a  stake  in  the  ground,  which  is 
85  feet  from  the  center  of  the  base  (the  ground  being 
supposed  to  be  level),  what  must  be  the  length  of 
the  rope  ? 

SOLUTION.— In  Fig.  29,  A  B  represents  the  stee- 
ple, 150  feet  high ;  C  a  stake  85  feet  from  the  foot 
of  the  steeple,  and  A  C  the  rope.  Here  we  have 
a  right  triangle  right-angled  at  B,  and  A  C  is  the 
hypotenuse.  The  square  of  A  B  —  ISO2  =22,500; 
of  C  />',  85*  =  7,225.  22,509  +  7,225  =  29,725 ;  4/29,725 
=  172.4  ft.,  nearly.  Ans. 

B  6O.      The    altitude 

of  any  triangle  is  a  line, 
as  B  D,  drawn  from  the 
vertex  B  of  the  -angle 
opposite  the  base  A  C, 
perpendicular  to  the 
base,  as  in  Fig.  30,  or 
to  the  base  extended,  as  in  Fig.  31. 

61.  If  in  any  parallelogram  a  straight  line,  called  the 
diagonal,  be  drawn,  connecting  two 
opposite  corners,  it  will  divide  the 
parallelogram  into  two  equal  triangles, 
as  A  D  B  and  D  B  C  in  Fig.  32.  The 
area  of  each  triangle  will  equal  one-half 
the  area  of  the  parallelogram,  i.  e.,  one-half  the  product  of  the 
base  and  the  altitude.  Hence,  to  find  the  area  of  any  triangle : 

Rule  8. — Multiply  the  base  by  the  altitude  and  divide  the 
product  by  2. 

EXAMPLE. — What  is  the  area  in  square  feet  of.  a  triangle  whose  base 
is  18  feet  and  whose  altitude  is  7  feet  9  inches  ? 

SOLUTION.—  7  ft.  9  in.  =  7f  ft.  =  -j-  ft,  18  X  -r  =  139|,  and  one- 
half  of  139£  =  69f  sq.  ft.  Ans. 

To  find  the  altitude  or  base  of  a  triangle,  having  given 
the  area  and  the  base  or  altitude: 

Rule  9.. — Multiply  the  area  by  2  and  divide  by  the  given 
dimension 


24  MENSURATION  AND  §  3 

EXAMPLE. — What  must  be  the  height  of  a  triangular  piece  of  sheet 
metal  to  contain  100  square  inches,  if  the  base  is  10  inches  long  ? 
SOLUTION.—    100  X  2  =  200 ;  200  -*-  10  =  20  in.     Ans. 

EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  area  of  a  triangle  whose  base  is  18  feet  long  and 
whose  altitude  is  10  feet  6  inches  ?  Ans.  94.5  sq.  ft. 

2.  Two  angles  of  a  scalene  triangle  together  equal  100°  4'.    What  is 
the  size  of  the  third  angle  ?  Ans.  79°  56'. 

3.  One  angle  of  a  right  triangle  equals  20°  10'  5".     What  is  the  size 
of  the  other  acute  angle  ?  Ans.  69°  49'  55". 

4.  A  ladder  65  feet  long  reaches  to  the  top  of  a  wall  when  its  foot 
is  25  feet  from  the  wall.     How  high  is  the  wall  ?  Ans.  60  ft. 

5.  Draw  a  triangle,  and  through  two  of  its  sides  draw  a  line  parallel 
to  the  base.     Letter  the  different  lines,  and  then,  without  referring  to 
the  text,  write  out  the  proportions  existing  between  the  sides  of  the 
two  triangles. 

6.  A  triangular  piece  of  sheet  metal  weighs  24  pounds.    If  the  base 
of  the  triangle  is  4  feet  and  its  height  6  feet,  how  much  does  the  metal 
weigh  per  square  foot  ?  Ans.   2  Ib. 

7.  The  area  of  a  triangle  is  16  square  inches.     If  the  altitude  is 
4  inches,  what  does  the  base  measure  ?  Ans.  8  in. 

8.  Two  sides  of  a  right  triangle  are  92  feet  and  69  feet  long.     How 
long  is  the  hypotenuse  ?  Ans.  115  ft. 

POLYGONS. 

62.  A  polygon  is  a  .plane  figure  bounded  by  straight 
lines.     The  term  is  usually  applied  to  a  figure  having  more 
than  four  sides.     The  bounding  lines  are  called  the  sides, 
and  the  sum  of  the    lengths   of  all  the  sides  is  called  the 
perimeter  of  the  polygon. 

63.  A  regular  polygon  is  one  in  which   all  the  sides 
and  all  the  angles  are  equal. 

64.  A  polygon  of  five  sides  is  called   a  pentagon  ;  one 
of  six  sides,  a  hexagon ;  one  of  seven  sides,  a  heptagon, 


Pentagon.       Hexagon.       Heptagon.       Octagon.          Decagon.        Dodecagon. 
FIG.  33. 

etc.     Regular  polygons  having  from  five  to  twelve  sides  are 


§  3 


USE  OF  LETTERS  IN  FORMULAS. 


25 


shown    in    Fig.    33. 
interior   angles,    as 


In    any  polygon,   the    sum   of  all  the 
A-\-B+C-\-D-\-E,    Fig.    34,  equals 


180°  multiplied  by  a  number  which  is  two 
less  than  the  number  of  sides  in  the  poly- 
gon. Hence,  to  find  the  size  of  any  one  of 
the  interior  angles  of  a  regular  polygon: 

Rule  1O.  —  Multiply  180°  by  the  num- 
ber of  sides  less  two  and  divide  the  result 
by  the  number  of  sides  ;  the  quotient 
will  be  the  number  of  degrees  in  each  interior  angle. 

EXAMPLE  1.—  If  Fig.  34  is  a  regular  pentagon,  how  many  degrees  are 
there  in  each  interior  angle  ? 

SOLUTION.  —  In  a  pentagon  there  are  five  sides;  hence,  5  —  2  =  3  and 
180  X  3  =  540  ;  540  -v-  5  =  108°  in  each  angle.  Ans. 


FIG.  si. 


EXAMPLE  2.— It  is  desired  to  make  a  miter-box 
in  which  to  cut  a  strip  of  molding  to  fit  around 
a  column  having  the  shape  of  a  regular  hexa- 
gon. At  what  angle  should  the  saw  run  across 
the  miter-box  ? 

SOLUTION.— In  Fig.  35,  let  A  B,  B  C,  CD, 
etc.  represent  the  pieces  of  molding  as  they 
will  fit  around  the  column.     First  find  the  size 
of.  one  of  the  equal  angles  of  the  polygon  by 
FIG.  35.  the  above   rule.      Number  of  sides  =  6;    6-2 

=  4;  hence,  180x4  =  720,  and  720^-6  =  120°  in  each  angle.  Now, 
let  M  N  represent  the  miter-box  and  O  S  the  direction  in  which  the 
saw  should  run ;  then,  A  B  O  is  the  angle  made  by  the  saw  with  the 
side  of  the  miter-box ;  but  as  the  polygon  is  a  regular  one,  this  angle 
is  one-half  the  interior  angle  ABC,  which  we  have  found  to  be  120°. 

120 
Hence,  the  saw  should  run  at  an  angle  of  -^-  =  60°  with  the  side  of  the 

miter-box.     Ans. 

65.  The  area  of  any  regular  polygon 
may  be  found  by  drawing  lines  from  the 
center  to  each  angle  and  computing  the 
area  of  each  triangle  thus  formed.  Hence, 
to  find  the  area  of  any  regular  polygon : 

Rule  11. — Multiply  the' length  of  a  side 
by  half  the  distance  from  the  side  to  the  FIG.  se. 

center,  and  that  product  by  the  number  of  sides.  The  last 
product  will  be  the  area  of  the  figure. 


26  MENSURATION  AND  §  3 

EXAMPLE. — In  Fig.  36  the  side  B  C  of  the  regular  hexagon  is 
12 inches  and  the  distance  A  O  is  10.4  inches;  required  the  area  of  the 
polygon. 

SOLUTION. —    10.4  -*-  2  =  5.2;  12  X  5.2  X  6  =  374.4  sq.  in.     Aris. 

66.  To  obtain  the  area  of 
any  irregular  polygon,  draw  diag- 
onals dividing  the  polygon  into 
triangles  and  quadrilaterals,  and 
compute  the  areas  of  these  sepa- 
rately ;  their  sum  will  be  the 
area  of  the  figure. 

EXAMPLE.— It  is  required  to  find  the  area  of  the  polygon  A  B  CD  EF, 
Fig.  37. 

SOLUTION.— Draw  the  diagonals  B  F  and  CF  and  the  line  FG 
perpendicular  to  D  E,  dividing  the  figure  into  the  triangles  A  B  F, 
B  C  F,  and  FG  E  and  the  rectangle  FC  D  G.  Let  it  be  supposed  that 
the  altitudes  of  the  figures  and  the  lengths  of  the  sides  A  B,  D  G,  and 
G  E  are  as  indicated  in  the  polygon  above.  Then, 

Area  A  B  F     =   16      7   =    56  sq.  in. 


Area  FC  D  G  =  14  X  10  =  140  sq.  in. 

Area  FG  E     =   9  *  10   =   45  sq.  in. 

Total  area  =  56  +  35  +  140  +  45  =  276  sq.  in.     Ans. 


EXAMPLES  FOR  PRACTICE. 

1.  How  many  degrees  are  there  in  one  of  the  angles  of  a  regular 
octagon?  Ans.   135°. 

2.  Find  the  area  of  the  polygon  A  B  CD  EF  (see  Fig.  37),  suppo- 
sing each  of  the  given  dimensions  to  be  increased  to  1^  times  the  length 
given  in  the  figure.  Ans.  621  sq.  in. 

3.  What  is  the  area  of  a  regular  heptagon  whose  sides  are  4  inches 
long,  the  distance  from  one  side  to  the  center  being  4.15  inches? 

Ans.  58.1  sq.  in. 

4.  At  what  angle  should  the  saw  run  in  a  miter-box  to  cut  strips 
to  fit  around  the  edge  of  a  table  top  made  in  the  shape  of  a  regular 
pentagon?  Ans.  54°. 


USE  OF  LETTERS  IN  FORMULAS. 


THE   CIRCLE. 

67.  A  circle  (Fig.  38)  is  a  figure  bounded 
by  a  curved  line,  called  the  circumference, 
every  point  of  which  is  equally  distant  from 
a  point  within,  called  the  center. 


11- 


-J* 


FIG.  39. 


68.  The  diameter  of  a  circle  is  a 
straight  line  passing  through  the  center 
and  terminated  at  both  ends  by  the  cir- 
cumference; thus,  A  B  (Fig.  39)  is  a  diam- 
eter of  the  circle. 


69.  The  radius  of  a  circle,  A  O  (Fig.  40), 
is  a  straight  line  drawn  from  the  center  O 
to  the  circumference.  It  is  equal  in  length 
to  one-half  the  diameter. 
;6  The  plural  of  radius  is  radii, 
and  all  radii  of  a  circle  are 
equal.  FIG.  40. 

7O.     An  arc  of  a  circle  (see  a  e  b,  Fig.  41) 
is  any  part  of  its  circumference. 

71.  A  chord  is  a  straight  line  joining 
any  two  points  in  a  circumference ;  or  it  is  a 
straight  line  joining  the  extremities  of  an 
arc ;  thus,  the  straight  line  A  J5,  Fig.  42,  is 
a  chord  of  the  circle  whose  corresponding 
arc  is  A  E  B. 


72.  An  inscribed  angle  is  one  whose 
vertex  lies  on  the  circumference  of  a 
circle  and  whose  sides  are  chords.  It 
is  measured  by  one-half  the  intercepted 
arc.  Thus,  in  Fig.  43,  A  B  C  is  an  in- 
scribed angle,  and  it  is  measured  by  one- 
half  the  arc  ADC. 


28  MENSURATION  AND  §  3 

EXAMPLE.— If  in  Fig.  43,  the  arc  A  D  C  =  f  of  the  circumference, 
what  is  the  measurement  of  the  inscribed  angle  A  B  C  ? 

SOLUTION. — Since  the  angle  is  an  inscribed  angle,  it  is  measured  by 
one-half  the  intercepted  arc,  or  f  X  1  =  \  of  the  circumference.  The 
whole  circumference  =  360° ;  hence,  360°  X  \  =  72° ;  therefore,  angle 
A  B  C  is  an  angle  of  72°. 

73.  If  a  circle  is  divided  into  halves,  each  half  is  called 
a   semicircle,    and    each    half    circumference    is    called    a 

semi-circumference. 

Any  angle  inscribed  in  a  semicircle  is 
a  right  angle,  since  it  is  measured  by 
}c  one-half  a  semi-circumference,  or  180° 
-=-  2  =  90°.  Thus,  the  angles  ADC 
and  A  B  C,  Fig.  44,  are  right  angles, 
since  they  are  inscribed  in  a  semicircle. 

74.  An  inscribed  polygon  is  one  whose  vertexes  lie  on 
the  circumference  of  a  circle  and  whose 

sides  are  chords,  as  A  B  C  D  E,  Fig.  45. 
The  sides  of  an  inscribed  regular  hex- 
agon have  the  same  length  as  the  radius 
of  the  circle. 

If,  in  any  circle,  a  radius  be  drawn 
perpendicular  to  any  chord,  it  bisects 
(cuts  in  halves)  the  chord.  Thus,  if  the  FIG.  45. 

radius   O  C,   Fig.   46,  is  perpendicular  to    the    chord  A  J3, 
A  D  =  DB. 

EXAMPLE.— If  a  regular  pentagon  is  inscribed 
in  a  circle  and  a  radius  is  drawn  perpendicular 
to  one  of  the  sides,  what  are  the  lengths  of  the 
two  parts  of  the  side,  the  perimeter  of  the  pen- 
tagon being  27  inches  ? 

SOLUTION.— A  pentagon  has  five  sides,  and" 
since  it  is  a  regular  pentagon,  all  the  sides  are 
of  equal  lengths ;  the  perimeter  of  the  pentagon, 
which  equals  the  distance  around  it,  or  equals 
the  sum  of  all  the  sides,  is  27  inches.  There- 
fore, the  length  of  one  side  =  27  -f-  5  =  5|  inches.  Since  the  penta- 
gon is  an  inscribed  pentagon,  its  sides  are  chords,  and  as  a  radius 
perpendicular  to  a  chord  bisects  it,  we  have  5|  -r-  2  =  2T7ff  inches,  which 
equals  the  length  of  each  of  the  parts  of  the  side  cut  by  a  radius  per- 
pendicular to  it.  Ans. 


§  3  USE  OF  LETTERS  IN  FORMULAS.  29 

75.  If,  from  any  point  on  the  circumference  of  a  circle, 
a  perpendicular  is  let  fall  upon  a  diameter,  it  will  divide 
the  diameter  into  two  parts,  one  of  ^^— —  „ 

which  will  be  in  the  same  ratio  to  the 
perpendicular  as  the  perpendicular  is  to 
the  other  part.     That  is,  the  perpendic-   A\ 
ular  will  be  a  mean  proportional  between 
the  two  parts  of  the  diameter. 

If  A  B,  Fig.  47,  is  the  given  diameter 
and  C  any  point  on  the  circumference,  FlG- 4~- 

then  AD:  CD=CD:DB,CD  being  a  mean  proportional 
between  A  D  and  D  B. 

EXAMPLE.— If  H K=  30  feet  and  IB  —  8  feet,  what  is  the  diameter 
of  the  circle,  H  K  being  perpendicular  to  A  B  ? 

SOLUTION.—  30  feet  -f-  2  feet  =  15  feet  =  I  H.  And  B  I  :  I H 
=  SH  :  I  A,  or  8:  15  =  15:  I  A. 

Therefore,   I A  =  ~  =  ~  =  28£  feet  and  IA  +  IB  =  28£  +  8 


feet  =  A  B,  the  diameter  of  the  circle.     Ans. 

76.  When  the  diameter  of  a  circle  and  the  lengths  of 
the  two  parts  into  which  it  is  divided  are  given,  the  length 
of  the  perpendicular  may  be  found  by  multiplying  the  lengths 
of  the  two  parts  together  and  extracting  the  square  root  of 
the  product. 

EXAMPLE.— In  Fig.  47,  the  diameter  of  the  circle  A  B  is  36£  feet 
and  the  distance  B  /is  8  feet;  what  is  the  length  of  the  line  H  K  1 

SOLUTION. — As  the  diameter  of  the  circle  is  36£  feet  and  as  B I  is 
8  feet,  I A  is  equal  to  36|  —  8  =  28£  feet.  The  two  parts,  therefore,  are 

8  and  28£  feet,  and  their  product  =  8  X  28£  =  8  X  ^  =  225 ;  the  square 

root  of  their  product  =  4/225  =  15  feet,  and  as  H  K  =  IH+  IK,  or 
2  Iff,  /fAT=  15X2  =  80  ft.     Ans. 

77.  To  find  the  circumference  of  a  circle,  the  diameter 
being  given: 

Rule  12.— Multiply  the  diameter  by  3.1416. 
EXAMPLE. — What  is  the  circumference  of  a  circle  whose  diameter  is 
15  inches  ? 

SOLUTION. —    15  X  3.1416  =  47.124  in.     Ans. 

78.  To  find  the  diameter  of  a  circle,  the  circumference 
being  given: 


30  MENSURATION  AND  §  3 

Rule  13. — Divide  the  circumference  by  3.1416. 
EXAMPLE. — What  is  the  diameter  of  a  circle  whose  circumference  is 
65. 973  inches? 

SOLUTION.—    65.973  -*-  3.1416  =  21  in.     Ans. 

79.  To  find  the  length  of  an  arc  of  a  circle : 

Rule  14. — Mtilttply  the  length  of  the  circumference  of  the 
circle  of  wJiich  the  arc  is  a  part  by  the  number  of  degrees  in 
the  arc  and  divide  by  360. 

EXAMPLE. — What  is  the  length  of  an  arc  of  24U,  the  radius  of  the  arc 
being  18  inches  ? 

SOLUTION.  —  18  X  2  =  36  in.  =  the  diameter  of  the  circle.  36 
X  3.1416  =  113.1  in.,  the  circumference  of  the  circle  of  which  the  arc  is 
a  part. 

24 
113.1  X  ogn  =  7-54  in->  or  the  length  of  the  arc.     Ans. 

80.  To  find  the  area  of  a  circle : 

Rule  15. — Square  the  diameter  and  multiply  by  .7854- 
EXAMPLE. — What  is  the  area  of  a  circle  whose  diameter  is  15  inches  ? 
SOLUTION.—     15*  =  225;  and  225  X  .7854  =  176.72  sq.  in.     Ans. 

81.  Given  the  area  of  a  circle,  to  find  its  diameter: 
Rule  16. — Divide  the  area  by  .  7854  and  extract  the  square 

root  of  the  quotient. 

EXAMPLE  1. — The  area  of  a  circle  =  17,671.5  square  inches.  What  is 
its  diameter  in  feet  ? 


SOLUTION.—    y     ^    =  150  inches. 

rp-j-  =  12^  feet,  or  the  diameter.     Ans. 

EXAMPLE  2. — What  is  the  area  of  a  flat  circular 
ring,  Fig.  48,  whose  outside  diameter  is  10  inches 
and  inside  diameter  is  4  inches  ? 

SOLUTION. — The  area  of  the  large  circle  =  10s 
X- 7854  =  78. 54  sq.  in.;  the  area  of  the  small 
circle  =  42  X  -7854  =  12.57  sq.  in.  The  area  of 
the  ring  is  the  difference  between  these  areas,  or 
FIG.  48.  78.54  —  12.57  =  65.97  sq.  in.  Ans. 

82.  To  find  the  area  of  a  sector  (a  sector  of  a  circle  is 
the  area  included  between  two  radii  and  the  circumference, 
as,  for  example,  the  area  B A  CO,  Fig.  36): 


§  3 


USE  OF  LETTERS  IN  FORMULAS. 


Rule  17.—  Divide  tJic  number  of  degrees  in  the  arc  of  the 
sector  by  360.  Multiply  the  result  by  the  area  of  the  circle  of 
i^Jiich  the  sector  is  a  part. 

EXAMPLE.  —  The  number  of  degrees  in  the  angle  formed  by  drawing 
radii  from  the  center  of  a  circle  to  the  extremities  of  the  arc  of  the 
circle  is  75°.  The  diameter  of  the  circle  is  12  inches;  what  is  the  area 
of  the  sector  ? 

SOLUTION.—    ^  =  A;  and  122  X  .7854  =  113.1  sq.  in. 
113.1  X  52  =  23.56  sq.  in.,  the  area.     Ans. 

83.  To  find  the  area  of  a  segment  of  a  circle  (a  seg- 
ment of  a  circle  is  the  area  included  between  a  chord  and 
its  arc;  for  example,  the  area  ABC,  Fig.  49)  when  its 
chord  and  height  are  given.  There  is  no  exact  method, 
except  by  applying  principles  of  trigonometry.  The  follow- 
ing rule  gives  results  that  are  exact  enough  for  practical 
purposes. 

Rule  18.  —  Divide  the  diameter  by  the  height  of  the  seg- 
ment ;  subtract  .608  from  the  quotient  and  extract  the  square 
root  of  the  remainder.  This  result  multiplied  by  4  times  the 
square  of  the  height  of  the  segment  and  then  divided  by  3 
will  give  the  area,  very  nearly. 

The  rule,  expressed  as  a  formula,  is  as  follows,  where 
D  =  the  diameter  of  the  circle  and  h  =  the  height  of  the 
segment  (see  Fig.  49)  : 


Area 


=-  *        -  .  608. 


EXAMPLE.—  What  is  the  area  of  the  segment 
of  a  circle  whose  diameter  is  54  inches,  the 
height  of  the  segment  being  20  inches  ? 

SOLUTION.—  Substituting  in  the  formula, 

~  4x400 


FIG.  49. 


Ans. 

NOTE.—  Had  the  chord  A  C,  Fig.  49,  been  given  instead  of  the 
diameter,  the  diameter  would  have  been  found  as  explained  in  Art.  75. 


//.  X.     J.—1Q 


32  MENSURATION  AND  §  3 

EXAMPLES  FOR  PRACTICE. 

1.  An  angle  inscribed  in  a  circle  intercepts  one-third  of  the  circum- 
ference.    How  many  degrees  are  there  in  the  angle  ?  Ans.  60°. 

2.  Suppose  tjiat  in   Fig.  47,  the  diameter  A  B  =  15  feet  and  the 
distance  B I  —  3  f eet.    What  is  the  length  of  the  line  H  K ?      Ans.  12  ft. 

3.  The  diameter  of  a  flywheel  is  18  feet.     What  is  the  distance 
around  it  to  the  nearest  16th  of  an  inch  ?  Ans.  56  ft.  6^  in. 

4.  A  carriage  wheel  was  observed  to  make  71f  turns  while  going 
300  yards.     What  was  its  diameter  ?  Ans.  4  ft. ,  nearly. 

5.  What  is  the  length  of  an  arc  of  04°,  the  radius  of  the  arc  being 
30  inches?  Ans.  33.51  in. 

6.  Find  the  area  of  a  circle  2  feet  3  inches  in  diameter. 

Ans.  3.976  sq.  ft. 

7.  What  must  be  the  diameter  of  a  circle  to  contain  100  square 
inches?  Ans.   11.28  in. 

8.  Compute  the  area  of  a  segment  whose  height  is  11  inches  and 
the  radius  of  whose  arc  is  21  inches.  Ans.  289.04  sq.  in. 

9.  Find  the  area  of  a  flat  circular  ring  whose  outside  diameter  is 
12  inches  and  whose  inside  diameter  is  6  inches.  Ans.  84.82  sq.  in. 


THE    PRISM    AND    CYLINDER. 

84.  A   solid,  or  body,  has   three   dimensions:    length, 
breadth,   and   thickness.     The    sides   which   enclose    it   are 
called  the  faces,  and  their  lines  of  intersection  are  called 
the  edges. 

85.  A  prism  is  a  solid  whose  ends  are  equal  and  par- 
allel polygons  and  whose  sides  are  parallelograms.      Prisms 
take  their  names  from  the  form  of  their  bases.     Thus,  a  tri- 
angular prism  is  one  having  a  triangle  for  its  base;  a  hex- 
agonal prism  is  one  having  a  hexagon  for  its  base,  etc. 

86.  A  cylinder  is  a  body  of'  uniform  diameter  whose 
ends  are  equal  parallel  circles. 

87.  A    parallelopipedoii     (Fig. 
50)  is  a  prism  whose  bases  (ends)  are 
parallelograms. 

88.  A  cube  (Fig.  51)  is  a  prism 
whose   faces   and    ends   are  squares. 

All  the  faces  of  a  cube  are  equal.  FlG-  51- 

In  the  case  of  plane  figures,  we  are  concerned 
with  perimeters  and  areas.     In  the  case  of  solids,  we  are 


§  3  USE  OF  LETTERS  IN  FORMULAS.  33 

concerned  with  the  areas  of  their  outside  surfaces  and  with 
their  contents  or  volumes. 

89.  The  entire  surface  of  any  solid  is  the  area  of  the 
whole  outside  of  the  solid,  including  the  ends. 

The  convex  surface  of  a  solid  is  the  same  as  the  entire 
surface,  except  that  the  areas  of  the  ends  are  not  included. 

90.  A  unit  of  volume  is  a  cube  each  of  whose  edges  is 
equal  in  length  to  the  unit.     The  volume  is  expressed  by 
the  number  of  times  it  will  contain  a  unit  of  volume. 

Thus,  if  the  unit  of  length  is  1  inch,  the  unit  of  volume 
will  be  the  cube  whose  edges  each  measure  1  inch,  this  cube 
being  1  cubic  inch  ;  and  the  number  of  cubic  inches  the  solid 
contains  will  be  its  volume.  If  the  unit  of  length  is  1  foot, 
the  unit  of  volume  will  be  1  cubic  foot,  etc.  Cubic  inch,  cubic 
foot,  and  cubic  yard  are  abbreviated  to  cu.  in.,  cu.  ft.,  and 
cu.  yd.,  respectively. 

Instead  of  the  word  volume,  the  expression  cubical  con- 
tents is  sometimes  used. 

91.  To  find  the  area  of  the  convex  surface  of  a  prism  or 
cylinder: 

Rule  19. — Multiply  the  perimeter  of  the  base  by  the  altitude. 

EXAMPLE  1.— A  block  of  marble  is  24  inches  long  and  its  ends  are 
9  inches  square.  What  is  the  area  of  its  convex  surface  ? 

SOLUTION.  —  9  X  4  =  36  =  the  perimeter  of  the  base;  36  X  34 
=  864  sq.  in.,  the  convex  area.  Ans. 

To  find  the  entire  area  of  the  outside  surface,  add  the  areas  of  the 
two  ends  to  the  convex  area.  Thus,  the  area  of  the  two  ends 
=  9  X  9  X  2  =  162  sq.  in. ;  864  +  162  =  1,026  sq.  in.  Ans. 

EXAMPLE  2. — How  many  square  feet  of  sheet  iron  will  be  required 
for  a  pipe  H  feet  in  diameter  and  10  feet  long,  neglecting  the  amount 
necessary  for  lapping  ? 

SOLUTION. — The  problem  is  to  find  the  convex  surface  of  a  cylinder 
1  £  feet  in  diameter  and  10  feet  long.  The  perimeter,  or  circumference, 
of  the  base  =  !$•  X  3.1416  =  1.5  X  3.1416  =  4.712  ft.  The  convex  sur- 
face =  4.712  X  10  =  47.12  sq.  ft.  of  metal.  Ans. 

92.  To  find  the  volume  of  a  prism  or  a  cylinder: 
Rule  2O. — Multiply  the  area  of  the  base  by  the  altitude. 
EXAMPLE  1. — What  is  the  weight  of  a  length  of  wrought-iron  shaft- 
ing 16   feet  long  and  2  inches  in  diameter  ?    Wrought  iron  weighs 
.28  pound  per  cubic  inch. 


MENSURATION  AND 


SOLUTION. — The  shaft  is  a  cylinder  16  ft.  long.  The  area  of  one 
end,  or  the  base,  =  2*  X  .7854  =  3.1416  sq.  in.  Since  the  weight  of 
the  iron  is  given  per  cubic  inch,  the  contents  of  the  shaft  must  be 
found  in  cubic  inches.  The  length,  16  ft.,  reduced  to  inches  =  16  X  12 
=  192  in.;  3.1416x192  =  603.19  cu.  in.  =  the  volume.  The  weight 
=  603. 19  X  .28  =  168.89  Ib.  Ans. 

EXAMPLE  2.— Find  the  cubical  contents  of  a  hexagonal  prism,  Fig.  52, 
12  inches  long,  each  edge  of  the  base  being  1  inch  long. 

SOLUTION. — In  order  to  obtain  the  area  of  one 
end,  the  distance  CD  from  the  center  Cto  one  side 
must  be  found. 

In  the  right  triangle  C  D  A,  side  A  D  =  $  A  B, 
or  |  inch,  and  since  the  polygon  is  a  hexagon, 
side  C  A  =  distance  A  B,  or  1  inch  (Art.  74). 
Hence,  C  A  being  the  hypotenuse,  the  length 
of  side  C  D  =  ^l*~—  (*)2  =  4/1*  -  .5*  =  4//T5,  or 
1  X 


FIG.  52. 


.866  inch.     Area  of  triangle  A  C  B  = 


=  .433  sq.  in. ;  area  of  the 


whole  polygon  =  .433  X  6  =  2.598  sq.  in.     Hence,  the  contents  of  the 
prism  =  2.598  X  12  =  31.176  cu.  in.     Ans. 

EXAMPLE  3. — It  is  required  to  find  the  number  of  cubic  feet  of  steam 
space  in  the  boiler  shown  in  Fig.  53.    The  boiler  is  16  feet  long  between 


FIG.  53. 

heads,  54  inches  in  diameter,  and  the  mean  water-line  M '  N  is  at  a 
distance  of  16  inches  from  the  top  of  the  boiler.  The  volume  of  the 
steam  outlet  casting  may  be  neglected. 

SOLUTION. — The  volume  of  the  steam  space,  which  is  that  space 
within  the  boiler  above  the  surface  M  NO  P  of  the  water,  is  found  by 
the  rule  for  finding  the  volume  of  a  prism  or  cylinder,  the  area  M  N  S 
being  the  base  and  the  length  NO  the  altitude.  First  obtain  the  area 
of  the  segment  M  N  S,  whose  height  //  is  16  inches,  in  square  feet;  then 
multiply  the  result  by  16,  the  length  of  the  boiler. 

By  the  formula  given  in  Art.  83,  the  area  of  the  segment  = 


§  3  USE  OF  LETTERS  IN  FORMULAS.  35 


27767  =  1.663. 

Hence,  the  area  =  341.33  X  1-663  =  567.63  sq.  in.  This  reduced  to 
sq.  ft.  =  567.63  -5-  144  =  3.942  sq.  ft.,  and  the  volume  therefore  =  3.942 
X  16  =  63.07  cu.  ft.  Ans. 

In  the  above  solution,  the  space  occupied  by  the  stays  is 
not  considered,  for  sake  of  simplicity.  They  are  not  shown 
in  the  figure. 

EXAMPLE  4. — In  the  above  boiler  there  are  60  tubes,  3J  inches  outside 
diameter.  How  many  gallons  of  water  will  it  take  to  fill  the  boiler  up 
to  the  mean  water  level,  there  being  231  cubic  inches  in  a  gallon  ? 

SOLUTION.— Find  the  volume  in  cubic  inches  of  that  part  of  the 
boiler  below  the  surface  of  the  water  M NO  P,  since  the  contents  of  a 
gallon  is  given  in  cubic  inches,  and  from  it  subtract  the  volume  of  the 
tubes  in  cubic  inches. 

This  may  be  done  by  first  finding  the  total  area  of  one  end  of  the 
boiler  in  square  inches,  from  it  subtracting  the  area  of  the  seg- 
ment M  N  S,  and  the  areas  of  the  ends  of  the  tubes  in  square  inches, 
and  then  by  multiplying  the  result  by  the  length  of  the  boiler  in  inches. 

Total  area  of  one  end  =  54*  X  -7854  =  2,290.23  sq.  in. 

Area  of  segment  M  N  S,  as  found  in  last  example,  =  567.63  sq.  in. 

Area  of  the  end  of  one  tube  =  8.25*  X  -7854  =  8.2958  sq.  in. 

Area  of  the  ends  of  the  60  tubes  =  8.2958  X  60  =  497.75  sq.  in. 

Hence,  the  area  to  be  subtracted  =  567.63  +  497.75  =  1,065.38  sq.  in. 
Subtracting,  2,290.23  -  1,065.38  =  1,224.85  sq.  in.  =  net  area. 

The  cubical  contents  =  1,224.85  X  16  X  12  =  235,171.2  cu.  in.  This 
divided  by  231  will  give  the  number  of  gallons;  whence,  235,171.2 
-T-  231  =  1,018.06  gal.  of  water.  Ans. 


EXAMPLES  FOR  PRACTICE. 

1.  Find  the  area  in  square  inches  of  the  convex  surface  of  a  bar  of 
iron  4J  inches  in  diameter  and  8  feet  5  inches  long.     Ans.  1,348.53  sq.  in. 

2.  Find  the  area  of  the  entire  surface  of  the  above  bar. 

Ans.  1,376.9  sq.  in. 

3.  What  is  the  area  of  the  entire  surface  of  the  hexagonal  prism 
whose  base  is  shown  in  Fig.  52  ?  Ans.  77.196  sq.,in. 

4.  A  multitubular  boiler  has  the  following  dimensions:  diameter, 
50  inches ;  length  between  heads,  15  feet ;  number  of  tubes,  56 ;  outside 
diameter  of  tubes,  3  inches ;  distance  of  mean  water-line  from  top  of 
boiler,  16  inches,     (a)  Compute  the  steam  space  in  cubic  feet,     (b)  Find 
the  number  of  gallons  of  water  required  to  fill  the  boiler  up  to  the  mean 
water-line.  .         ( (a)    56.4  cu.  ft. 

\(b)    800  gal. 


36 


MENSURATION  AND 


THE  PYRAMID  AND  CONE. 

93.     A   pyramid   (Fig.    54)   is   a    solid  whose  base  is  a 
polygon    and  whose  sides  are 
triangles  uniting  at  a  common 
point,  called  the  vertex. 

94.  A  cone  (Fig.  55)  is  a 
solid  whose  base  is  a  circle  and 
whose  convex  surface  tapers 
uniformly  to  a  point  called  FIG.  55. 


FIG.  54. 
the  vertex. 

95.  The  altitude  of  a  pyramid  or  cone  is  the  perpen- 
dicular distance  from  the  vertex  to  the  base. 

96.  The  slant  height  of  a  pyramid  is  a  line  drawn  from 
the  vertex  perpendicular  to  one  of  the  sides  of  the   base. 
The  slant  height  of  a  cone  is  any  straight  line  drawn  from 
the  vertex  to  the  circumference  of  the  base. 

97.  To  find  the  convex  area  of  a  pyramid  or  cone  : 
Rule  21.  —  Multiply  the  perimeter  of  tJie  base  by  one-Jialf 

the  slant  height. 

EXAMPLE  1.  —  What  is  the  convex  area  of  a  pentagonal  pyramid  if 
one  side  of  the  base  measures  6  inches  and  the  slant  height  is  14  inches  ? 

SOLUTION.  —  The  base  of  a  pentagonal  pyramid  is  a  pentagon,  and, 
consequently,  has  fives  sides. 

6  X  5  =  30  inches,  or  the  perimeter  of  the  base.  30  X  -«  =  210  sq.  in., 
or  the  convex  area.  Ans. 

EXAMPLE  2.—  What  is  the  entire  area  of  a  right  cone  whose  slant 
height  is  17  inches  and  whose  base  is  8  inches  in  diameter  ? 

SOLUTION.  —  The  perimeter  of  the  base  =  8  X  3.1416  =  25.1328  in. 
25.1328  X       =  213.63  sq.  in. 


Convex  area 
Area  of  base  = 


8s  X  .7854  =    50.27  sq.  in. 


Entire  area  =  263.90  sq.  in.     Ans. 
98.     To  find  the  volume  of  a  pyramid  or  cone: 
Rule  22.  —  Multiply  the  area  of  the  base  by  one-third  of 

the  altitude. 

EXAMPLE  1.  —  What  is  the  volume  of  a  triangular  pyramid,  each  edge 

of  whose  base  measures  6  inches  and  whose  altitude  is  8  inches  ? 


USE  OF  LETTERS  IN  FORMULAS. 


37 


SOLUTION. — Draw  the  base  as  shown  in  Fig.  56; 
it  will  be  an  equilateral  triangle,  all  of  whose  sides 
are  6  inches  long. 

Draw  a  perpendicular  B  D  from  the  vertex  to 
the  base;  it  will  divide  the  base  into  two  equal 
parts,  since  an  equilateral  triangle  is  also  isosceles, 
and  will  be  the  altitude  of  the  triangle.  In  order 
to  obtain  the  area  of  the  base,  this  altitude  must 
be  determined. 

In  the  right  triangle  B  D  A,  the  hypotenuse  B  A  =  6  inches  and 
side  A  D  =  3  inches,  to  find  the  other  side, 

B  D  =  |/6*2  —  32  =  5.2  in.,  nearly. 
6X5.2 


Area  of  the  base,  or  B  A  C,  = 

Q 

ume  =  15.6  X  o  =  41.6  cu.  in.     Ans. 


15.6  sq.  in.    Hence,  the  vol- 


Ex  AMPLE   2.—  What    is   the   volume  of  a   cone  whose  altitude    is 
18  inches  and  whose  base  is  14  inches  in  diameter  ? 

SOLUTION.—  Area  of  the  base  =  142  X  .7854  =  153.94  sq.  in.     Hence, 


1  8 

the  volume  =  153.94  X  -5-  =  923.64  cu.  in. 
o 


Ans. 


EXAMPLES  EOB  PRACTICE. 

1.  Find  the  convex  surface  of  a  square  pyramid  whose  slant  height 
is  28  inches  and  one  edge  of  whose  base  is  7£  inches  long. 

Ans.  420  sq.  in. 

2.  What  is  the  volume  of  a  triangular  pyramid,  one  edge  of  whose 
base  measures  3  inches  and  whose  altitude  is  4  inches  ?      Ans.  5.2cu.in. 

3.  Find  the  volume  of  a  cone  whose  altitude  is  12  inches  and  the 
circumference  of  whose  base  is  31.416  inches.  Ans.  314.16  cu.  in. 

NOTE. — Find  the  diameter  of  the  base  and  then  its  area. 


THE  FRUSTUM  OF  A  PYRAMID  OB  CONE. 

99.  If  a  pyramid  be  cut  by  a 
plane,  parallel  to  the  base,  so  as  to 
form  two  parts,  as  in  Fig.  57,  the 
lower  part  is  called  the  frustum 
of  the  pyramid. 

If  a  cone  be  cut  in  a  similar  man- 
ner, as  in  Fig.  58,  the  lower  part  is 
called  the  frustum,  of  the  cone. 


FIG.  57. 


FIG.  58. 


38  MENSURATION  AND  §  3 

100.  The  upper  end  of  the  frustum  of  a   pyramid  or 
cone  is  called  the  tipper  base,  and  the  lower  end  the  lower 
"base.     The  altitude  of  a  frustum  is  the  perpendicular  dis- 
tance between  the  bases. 

101.  To   find    the    convex    surface   of    a  frustum  of    a 
pyramid  or  cone: 

Rule  23.  —  Multiply  one-half  the  sum  of  the  perimeters  of 
the  t^vo  bases  by  the  slant  height  of  tlie  frustum. 

EXAMPLE  1.  —  Given,  the  frustum  of  a  triangular  pyramid,  in  which 
one  side  of  the  lower  base  measures  10  inches,  one  side  of  the  upper  ' 
base  measures  6  inches,  and  whose  slant  height  is  9  inches  ;  find  the 
area  of  the  convex  surface. 

SOLUTION.  —    10  in.  X  3  =  30  in.,  the  perimeter  of  the  lower  base. 

6  in.  X  3  =  18  in.,  the  perimeter  of  the  upper  base. 

OA      I      -J  Q 

—  =  24  in.,  or  one-half  the  sum  of  the  perimeters  of  the  two 
bases.  24  X  9  =  216  sq.  in.,  the  convex  area.  Ans. 

EXAMPLE  2.  —  If  the  diameters  of  the  two  bases  of  a  frustum  of  a 
cone  are  12  inches  and  8  inches,  respectively,  and  the  slant  height  is 
12  inches,  what  is  the  entire  area  of  the  frustum  ? 


SOLUTION.-  -  xia  =  876j99  ^    ,        the 

area  of  the  convex  surface. 

Area  of  the  upper  base  =    8s  X  -7854  =  50.27  sq.  in. 
Area  of  the  lower  base  =  12*  X  .7854  =  113.1  sq.  in.  • 
The  entire  area  of  the  frustum  =  376.99  -f-  50.27  •+•  113.1  =  540.36  sq.  in. 

Ans. 

1O2.  To  find  the  volume  of  the  frustum  of  a  pyramid 
or  cone  : 

Rule  24.  —  Add  togetJier  the  areas  of  the  upper  and  lower 
bases  and  the  square  root  of  the  product  of  the  two  areas  ; 
multiply  the  sum  by  one-third  of  the  altitude. 

EXAMPLE  1.  —  Given,  a  frustum  of  a  square  pyramid  (one  whose  base 
is  a  square);  each  edge  of  the  lower  base  is  12  inches,  each  edge  of  the 
upper  base  is  5  inches,  and  its  altitude  is  16  inches  ;  what  is  its  volume? 

SOLUTION.—  Area  of  upper  base  =  5  X  5  =  25  sq.  in.  ;  area  of  lower 
base  =  12  X  12  =  144  sq.  in.  ;  the  square  root  of  the  product  of  the  two 
areas  =  |/25  X  144  =  60.  Adding  these  three  results,  and  multiplying 

by  one-third  the  altitude,  25  +144  +  60  =  229;  229  X  ^  =  1.221$  cu.  in, 
=  the  volume.     Ans. 


§  3  USE  OF  LETTERS  IN  FORMULAS.  39 

EXAMPLE  2. — How  many  gallons  of  water  will  a  round  tank  hold 
which  is  4  feet  in  diameter  at  the  top,  5  feet  in  diameter  at  the  bottom, 
and  8  feet  deep  ? 

SOLUTION.— There  are  231  cubic  inches  in  a  gallon,  and  the  volume 
of  the  tank  should  be  found  in  cubic  inches.  The  tank  is  in  the  shape 
of  the  frustum  of  a  cone.  The  upper  diameter  =  4  X  12  =  48  inches ; 
the  lower  diameter  =  5  X  12  =  60  inches,  and  the  depth  =  8  X  12 

—  96  inches.     Area  of  upper  base  =  48s  X  .7854=  1,809.56  sq.  in. ;  area 
of   lower   base  =  60'2  X  .7854  =  2,827.44  sq.    in. ;    -/1, 809. 56^X^7827^4 
=  2,261.95. 

Qfi 

Whence,    1,809.56  +  2,827.44  +  2,261.95  =  6,898.95;    6,898.95  X  y 

—  220,766.4  cu.  in.  =  contents.     Now,  since  there  are  231  cu.  in.  in 
1  gallon,  the  tank  will  hold  220,766.4  -=-  231  =  955.7  gal.,  nearly.      Ans. 


EXAMPLES  FOR  PRACTICE. 

1.  Find  the  convex  surface  of  the  frustum  of  a  square  pyramid,  one 
edge  of  whose  low.er  base  is  15  inches  long,  one  edge  of  whose  upper 
base  is  14  inches  long,  and  whose  slant  height  is  1  inch.    Ans.  58  sq.  in. 

2.  Find  the  volume  of  the  above  frustum,  supposing  its  altitude  to 
be  3  inches.  Ans.  631  cu.  in. 

3.  Find  the  volume  of  the  frustum  of  a  cone  whose  altitude  is  12  feet 
and  the  diameters  of  whose  upper  and  lower  bases  are  8  and  10  feet, 
respectively.  Ans.  766.55  cu.  ft. 

4.  If  a  tank  had  the  dimensions  of  example  3,  how  many  gallons 
would  it  hold?  Ans.  5,734.2  gal.,  nearly. 


THE   SPHERE   AND   CYLINDRICAL  RING. 

103.  A  sphere  (Fig.  59)  is  a  solid  bounded 
by  a  uniformly  curved  surface,  every  point 
of    which    is   equally   distant    from    a   point 
within,  called  the  center. 

The  word  ball,  or  globe,  is  generally  used 
instead  of  sphere. 

104.  To  find  the  area  of  the  surface  of  a          FIG  59 
sphere : 

Rule  25. — Square  the  diameter  and  multiply  the  result 
by  3. 1416. 

EXAMPLE. — What  is  the  area  of  the  surface  of  a  sphere  whose  diam- 
eter is  14  inches  ? 

SOLUTION.  —  Diameter  squared  X  3.1416  =  14J  X  3.1416  =  14  X  14 
X  3.1416  =  615.75  sq.  in.  Ans. 


40  MENSURATION  AND  §  3 

From  this  it  will  be  seen  that  the  surface  of  a  sphere 
equals  the  circumference  of  a  great  circle  multiplied  by  the 
diameter,  a  rule  often  used ;  a  great  circle  of  a  sphere  is  the 
intersection  of  its  surface  with  a  plane  passing  through  its 
center ;  for  instance,  the  great  circle  of  a  sphere  6  inches 
diameter  is  a  circle  of  6  inches  diameter.  Any  number  of 
great  circles  could  be  described  on  a  given  sphere. 

1O5.     To  find  the  volume  of  a  sphere : 

Rule  26. — Cube  the  diameter  and  multiply  the  result  by 


EXAMPLE. — What  is  the  weight  of  a  lead  ball  12  inches  in  diameter, 
a  cubic  inch  of  lead  weighing  .41  pound  ? 

SOLUTION. — Diameter  cubed  X  .5236  =  12  X  12  X  12  X  .5236  =  904.78 
cu.  in.,  or  the  volume  of  the  ball.  The  weight,  therefore,  =904.78 
X  .41=370.961b.  Ans. 

1O6.     To  find  the  convex  area  of  a  cylindrical  ring: 

A  cylindrical  ring  (Fig.  60)  is  a  cyl- 
inder bent  to  a  circle.     The  altitude  of 
the  cylinder  before  bending  is  the  same 
\B    as  the  length  of  the  dotted  center  line  D. 
The   "base    will   correspond    to    a   cross- 
section  on  the  line  A  B  drawn  from  the 
center    O.     Hence,   to    find  the  convex 
FIG.  eo.  area : 

Rule  27. — Multiply  the  circumference  of  an  imaginary 
cross-section  on  the  line  A  B  by  the  length  of  the  center 
line  D. 

EXAMPLE. — If  the  outside  diameter  of  the  ring  is  12  inches  and  the 
inside  diameter  is  8  inches,  what  is  its  convex  area  ? 

SOLUTION. — The  diameter  of  the  center  circle  equals  one-half  the  sum 

-|  e)    .     Q 

of  the  inside  and  outside  diameters  =  — 5— =  10,   and   10x3.1416 

=  31.416  in.,  the  length  of  the  center  line. 

The  radius  of  the  inner  circle  is  4  inches;  of  the  outside  circle, 
6  inches ;  therefore,  the  diameter  of  the  cross-section  on  the  line  A  B 
is  2  inches.  Then,  2  X  3.1416  =  6.2832  in.,  and  6.2832  X  31.416 
=  197.4  sq.  in.,  the  convex  area.  Ans. 


§  3  USE  OF  LETTERS  IN  FORMULAS.  41 

1O7.     To  find  the  volume  of  a  cylindrical  ring: 

Rule  28. —  TJic  volume  will  be  tlie  same  as  that  of  a  cylin- 
der whose  altitude  equals  the  length  of  the 
dotted  center  line  D  (Fig.  61)  and  whose 
base  is  the  same  as  a  cross-section  of  the 
ring  on  the  line  A  B  drawn  from  the 
center  O.  Hence,  to  find  the  volume  of  a 
cylindrical  ring,  multiply  the  area  of  an 
imaginary  cross-section  on  the  line  A  B 
by  the  length  of  the  center  line  D.  FlG-  01- 

EXAMPLE. — What  is  the  volume  of  a  cylindrical  ring  whose  outside 
diameter  is  12  inches  and  whose  inside  diameter  is  8  inches? 

SOLUTION. — The  diameter  of  the  center  circle  equals  one-half  the 
sum  of  the  inside  and  outside  diameters  =  -^— — •  =  10. 

10  X  3.1416  =  31.416  inches,  the  length  of  the  center  line. 

The  radius  of  the  outside  circle  =  6  inches;  of  the  inside  circle 
=  4  inches ;  therefore,  the  diameter  of  the  cross-section  on  the  line  A  B 
=  2  inches. 

Then,  22  X  .7854  =  3.1416  sq.  in.,  the  area  of  the  imaginary  cross- 
section. 

And  3.1416  X  31.416  =  98.7  cu.  in.,  the  volume.     Ans. 


EXAMPLES  FOR  PRACTICE. 

1.  What  is  the  volume  of  a  sphere  30  inches  in  diameter  ? 

Ans.  14, 137. 2  cu.  in 

2.  How  many  square  inches  in  the  surface  of  the  above  sphere  ? 

Ans.  2,827.44  sq.  ia 

3.  Required  the  area  of  the  convex  surface  of  a  circular  ring,  the 
outside  diameter  of  the  ring  being  10  inches  and  the  inside  diameter 
7fc  inches.  Ans.  107.95  sq.  in. 

4.  Find  the  cubical  contents  of  the  ring  in  the  last  example. 

Ans.  33.73  cu.  in. 

5.  The  surface  of  a  sphere  contains  314.16  square  inches.     What  K 
the  volume  of  the  sphere  ?  Ans.  523.6  cu.  in. 


PRINCIPLES  OF  MECHANICS. 


MATTER  AOT)  ITS  PROPERTIES. 


DEFINITION  OF  MECHANICS. 

1.  Mechanics  is  that  science  which  treats  of  the  action 
of  forces  on  bodies  and  the  effects  that  they  produce ;  it 
treats  of  the  laws  that  govern  the  movement  and  equilibrium 
of  bodies  and  shows  how  they  may  be  applied. 


MATTER. 

2.  Matter  is  anything  that  occupies  space.     It  is  the 
substance  of  which  all  bodies  consist.      Matter  is  composed 
of  molecules  and  atoms. 

3.  A  molecule  is  the  smallest  portion  of  matter  that 
can  exist  without  changing  its  nature. 

4.  An  atom  is  an  indivisible  portion  of  matter. 
Atoms  unite  to  form  molecules,  and  a  collection  of  mole- 
cules forms  a  mass  or  body. 

A  drop  of  water  may  be  divided  and  subdivided  until 
each  particle  is  so  small  that  it  can  only  be  seen  by  the  moat 
powerful  microscope,  but  each  particle  will  still  be  water. 

Now,  imagine  the  division  to  be  carried  on  still  further, 
until  a  limit  is  reached  beyond  which  it  is  impossible  to  go 
without  changing  the  nature  of  the  particle.  The  particle 
of  water  is  now  so  small  that,  if  it  be  divided  again,  it  will 

For  notice  of  the  copyright,  see  page  immediately  following  the  title  page. 


2  PRINCIPLES  OF  MECHANICS.  §  4 

cease  to  be  water,  and  will  be  something  else;  we  call  this 
particle  a  molecule. 

If  a  molecule  of  water  be  divided,  it  will  yield  2  atoms  of 
hydrogen  gas  and  1  of  oxygen  gas.  If  a  molecule  of  sul- 
phuric acid  be  divided,  it  will  yield  2  atoms  of  hydrogen,  1  of 
sulphur,  and  4  of  oxygen. 

5.  Bodies   are   composed    of    collections   of    molecules. 
Matter  exists  in  three  conditions  or  forms :  solid,  liquid,  and 
gaseous. 

6.  A  solid  "body  is  one  whose  molecules  change  their 
relative    positions   with    great    difficulty;    as    iron,    wood, 
stone,  etc. 

7.  A  liquid  "body  is  one  whose  molecules  tend  to  change 
their  relative  positions  easily.     Liquids  readily  adapt  them- 
selves to  the  shape  of  vessels  that  contain  them,  and  their 
upper  surface  always  tends  to  become  perfectly  level.    Water, 
mercury,  molasses,  etc.,  are  liquids. 

8.  A  gaseous  body,  or  gas,  is  one  whose  molecules  tend 
to  separate  from  one  another;  as  air,  oxygen,  hydrogen,  etc. 

Gaseous  bodies  are  sometimes  called  aeriform,  (air-like) 
bodies.  They  are  divided  into  two  classes:  the  so-called 
"permanent"  gases  and  vapors. 

9.  A  permanent  gas  is  one  that  remains  a  gas  at  ordi- 
nary temperatures  and  pressures. 

10.  A  vapor  is  a  body  that  at  ordinary  temperatures  is  a 
liquid  or  solid,  but  when  heat  is  applied,  becomes  a  gas,  as 
steam. 

One  body  may,  under  different  conditions,  exist  in  all 
three  states;  as,  for  example,  mercury,  which  at  ordinary 
temperatures  is  a  liquid,  becomes  a  solid  (freezes)  at  40°  below 
zero,  and  a  vapor  (gas)  at  600°  above  zero.  By  means  of 
great  cold,  all  gases,  even  hydrogen,  have  been  liquefied, 
and  many  solidified. 

By  means  of  heat,  all  solids  have  been  liquefied,  and  a 
great  many  vaporized.  It  is  probable  that,  if  we  had  the 
means  of  producing  sufficiently  great  extremes  of  heat  and 


§  4  PRINCIPLES  OF  MECHANICS.  3 

cold,  all  solids  might  be  converted  into  gases,  and  all  gases 
into  solids. 

1 1.  Every  portion  of  matter   possesses  certain  qualities 
called  properties.     Properties  of  matter  are  divided  into  two 
classes :  general  and  special. 

12.  General  properties  of  matter  are  those  that  are 
common   to  all  bodies.     They  are  as  follows:     Extension, 
impenetrability,   weight,   indestructibility,    inertia,    mobility, 
divisibility,     porosity,     compressibility,      expansibility,     and 
elasticity, 

13.  Extension  is  the  property  of  occupying  space.     Since 
all  bodies  must  occupy  space,  it  follows  that   extension  is  a 
general  property. 

14.  By  Impenetrability  we  mean  that  no  two  bodies 
can  occupy  exactly  the  same  space  at  the  same  time. 

15.  Weight  is  the    measure  of   the    earth's    attraction 
upon  a  body.     All  bodies  have  weight.     In  former  times  it 
was  supposed  that  gases  had  no  weight,  since,  if  unconfmed, 
they  tend  to  move  away  from  the  earth,  but,  nevertheless, 
they  will  finally  reach  a  point  beyond  which  they  cannot  go, 
being  held  in  suspension  by  the  earth's  attraction.     Weight 
is  measured  by  comparison  with  a  standard.     The  standard 
is  a  bar  of  platinum  weighing  1  pound,   owned  and  kept  by 
the  Government. 

16.  Inertia   means   that    a   body   cannot    put  itself   in 
motion  nor  bring  itself  to  rest.     To  do  either  it  must  be 
acted  upon  by  some  force. 

17.  Mobility  means   that    a   body  can    be    changed   in 
position  by  some  force  acting  upon  it. 

18.  Divisibility  is  that  property  of  matter  by  virtue^  of 
which  a  body  may  be  separated  into  parts. 

19.  Porosity  is  the  term  used  to  denote  the  fact   that 
there  is  space  between  the  molecules  of  a  body.     The  mole- 
cules of  a  body  are  supposed  to  be  spherical,   and,  hence, 
there  is  space  between  them,   as  there  would  be  between 


4  PRINCIPLES  OF  MECHANICS.  §  4 

peaches  in  a  basket.  The  molecules  of  water  are  larger 
than  those  of  salt ;  so  that  when  salt  is  dissolved  in  water, 
its  molecules  wedge  themselves  between  the  molecules  of  the 
water,  and,  unless  too  much  salt  is  added,  the  water  will 
occupy  no  more  space  than  it  did  before.  This  does  not 
prove  that  water  is  penetrable,  for  the  molecules  of  salt 
occupy  the  space  that  the  molecules  of  water  did  not. 

Water   has  been  forced  through  iron  by  pressure,    thus 
proving  that  iron  is  porous. 

20.  Compressibility  is  a  natural  consequence  of  poros- 
ity.    Since  there  is  space  between  the  molecules,  it  is  evi- 
dent that  by  means  of  force  (pressure)  they  can  be  brought 
closer  together,    and  thus  the  body  be  made  to  occupy  a 
smaller  space. 

21.  Expansibility  is  the  term  used  to  denote  the  fact 
that  the  molecules  of  a  body  will,  under  certain  conditions 
(when  heated,   for    example),   move    farther  apart,  and   so 
cause  the  body  to  expand,  or  occupy  a  greater  space. 

22.  Elasticity  is  that  property  of  matter  which  enables 
a  body  when  distorted  within  certain  limits  to  resume  its 
original  form  when  the  distorting  force  is  removed.     Glass, 
ivory,   and  steel  are  very  elastic,   clay  and  putty  in  their 
natural  state  being  very  slightly  so. 

23.  Indestructibility  is  the  term  used  to  denote  the 
fact  that  we  cannot  destroy  matter.     A  body  may  undergo 
thousands  of  changes,  be  resolved  into  its  molecules,  and 
its  molecules  into  atoms,  which  may  unite  with  other  atoms 
to    form   other   molecules  and  bodies   entirely  different    in 
appearance  and  properties  from  the  original  body,  but  the 
same    number   of   atoms   remain.      The   whole    number   of 
atoms  in  the  universe  is  exactly  the  same  now  as  it  was 
millions  of  years  ago,  and  will  always  be  the  same.     Matter 
is  indestructible. 

24.  Special   properties   are    those   that    are   not   pos- 
sessed by  all  bodies.     Some  of  the  most  important  are  as 


§  4  PRINCIPLES  OF  MECHANICS.  5 

follows:    hardness,    tenacity,    brittleness,    malleability,     and 
ductility. 

25.  Hardness. — A  piece  of  copper  will  scratch  a  piece 
of  wood,  steel  will  scratch  copper,  and  tempered  steel  will 
scratch  steel  in  its  ordinary  state.     We  express  all  this  by 
saying  that  steel  is  harder  than  copper,  and  so  on.     Emery 
and  corundum  are  extremely  hard,  and  the  diamond  is  the 
hardest  of  all  known  substances.     It   can  only  be  polished 
with  its  own  powder. 

26.  Tenacity  is  the  term    applied   to   the   power  with 
which  some  bodies  resist  a  force  tending  to  pull  them  apart. 
Steel  is  very  tenacious. 

27.  Brittleness.  —  Some    bodies    possess    considerable 
power  to  resist  either  a  pull  or  a  pressure,   but  they  are 
easily  broken  when  subjected  to  shocks  or  jars;  for  exam- 
ple, good  glass  will  bear  a  greater  compressive  force  than 
most  woods,  but  may  be  easily  broken  when  dropped  upon 
a  hard  floor;  this  property  is  called  brittleness. 

28.  Malleability  is  that  property  which  permits  of  some 
bodies  being  hammered  or  rolled  into  sheets.     Gold  is  the 
most  malleable  of  all  substances. 

29.  Ductility   is    that    property    which    enables   some 
bodies  to  be  drawn  into  wire.     Platinum  is  the  most  ductile 
of  all  substances. 


MOTION  AOT)  VELOCITY. 


DEFINITIONS. 

3O.  Motion  is  the  opposite  of  rest  and  indicates  a  chang- 
ing of  position  in  relation  to  some  object  which  is  for  that 
purpose  regarded  as  being  fixe'd.  If  a  large  stone  is  rolled 
down  hill,  it  is  in  motion  in  relation  to  the  hill. 

If  a  person  is  on  a  railway  train  and  walks  in  the  oppo- 
site direction  from  that  in  which  the  train  is  moving,  and 
with  the  same  speed,  he  will  be  in  motion  as  regards  the 

H.  8.    I.— 11 


6  PRINCIPLES  OF  MECHANICS.  §  4 

train,  but  at  rest  with  respect  to  the  earth,  since,  until  he 
gets  to  the  end  of  the  train,  he  will  be  directly  over  the 
spot  at  which  he  was  when  he  started  to  walk. 

31.  The  path,  of  a  body  In  motion  is  the  line  described 
by  a  certain  point  in  the  body  called  its  center  of  gravity. 
No  matter  how  irregular  the  shape  of  the  body  may  be,  nor 
how  many  turns  and  twists  it  may  make,  the  line  that  indi- 
cates the  direction  of  this  point  for  every  instant  that  it  is 
in  motion  is  the  path  of  the  body. 

32.  Velocity  is  rate  of  motion.     It   is  measured  by  a 
unit  of  space  passed  over  in  a  unit  of  time.     When  equal 
spaces  are  passed  over  in  equal  times,  the  velocity  is  said 
to  be  uniform. 

If  the  flywheel  of  an  engine  keeps  up  a  constant  speed  of 
a  certain  number  of  revolutions  per  minute,  the  velocity  of 
any  point  is  uniform.  A  railway  train  having  a  constant 
speed  of  40  miles  per  hour  moves  40  miles  every  hour, 
or  £-§-  =  !•  mile  every  minute,  and  since  equal  spaces  are 
passed  over  in  equal  times,  the  velocity  is  uniform. 

33.  Variable  Velocity. — When  a  body  moves  in  such 
a  way  that  the  spaces  passed  over  in  equal  periods  of  time 
are  unequal,  its  velocity  is  said  to  be  variable. 

34.  The  rate  of  motion  of  a  body  with  a  variable  veloc- 
ity may  increase  or  decrease  at  a  uniform  rate.     When  the 
velocity  varies  in  either  of  these  ways,  the  body  is  said  to 
have  a  uniformly  varying  velocity. 

The  most  familiar  example  of  uniformly  varying  velocity 
is  a  falling  weight.  Suppose  a  stone  is  dropped  from  a  high 
bridge.  It  starts  from  a  state  of  rest  with  no  velocity,  but 
its  velocity  constantly  increases  until  it  strikes.  Its  increase 
in  velocity  during  any  equal  units  of  time  is  nearly  the 
same.  Thus,  at  the  end  of  the  first  second  its  velocity  will 
have  increased  from  0  to  a  rate  of  32.16  feet  per  second, 
nearly;  during  the  next  second  the  velocity  will  have 
increased  by  the  same  amount,  making  the  velocity  at  the 
end  of  the  second  second  64. 32  feet.  At  the  end  of  the  third 


§  4  PRINCIPLES  OF  MECHANICS.  7 

second  a  like  increase  will  have  taken  place,  and  the  veloc- 
ity will  then  be  3  X  32.16  =  96.48  feet  per  second. 

35.  The  change  in  the  velocity  of  a  body  during  a  period 
of  time  is  called  its  acceleration  for  that  period.     Thus,  in 
the  case  of  the  falling  weight  just  considered,  the  change 
of    32.16  feet   per  second  that  takes  place  in   its  velocity 
during  each  second  is  its  acceleration  in  feet  per  second  for 
each  second  considered.      If  the  period  of  time  considered 
had  been  2  seconds,  the  acceleration  would  have  been  the 
increase  in  velocity  during  this  time,  that  is,  64.32  feet  per 
second  for  each  2  seconds  considered. 

36.  The  change  in  velocity  may  be  from  a  higher  to  a 
lower  rate.    Thus,  when  a  stone  is  thrown  upwards,  it  leaves 
the  hand  with  a  given  velocity;  its  upward  motion  is  con- 
stantly resisted  by  the  force  of  gravity  and  the  resistance 
of  the  air,  and  in  consequence  of  these  resistances,  it  moves 
slower  and  slower  until  it  finally  stops  and  begins  to  return 
to  the  earth.     A  change  in  velocity  of  this  kind  is  some- 
times called  retardation. 

37.  The  mean  or  average  velocity  of  a  body  moving 
with  a  variable  velocity  can  only  be  given  for  a  stated  period 
of  time,  and  is  numerically  equal  to  the  uniform  velocity 
that  will  take  the  body  over  the  same  distance  in  the  same 
time. 


RULES  FOR  VELOCITY  PROBLEMS. 

38.     Uniform  and  Average  Velocity. — 

Let  d=  distance; 

/  =  time; 
v  =  velocity. 

Riile  1. —  To  find  the  uniform  or  the  average  velocity  that 
a  body  must  have  to  pass  over  a  certain  distance  or  space  in  a 
given  time,  divide  the  distance  by  the  time. 

Or,  v=d7. 


8  PRINCIPLES  OF  MECHANICS.  §  4 

EXAMPLE  1. — The  piston  of  a  steam  engine  travels  3,000  feet  in 
5  minutes  ;  what  is  its  average  velocity  in  feet  per  minute  ? 

SOLUTION. — Here  3,000  feet  is  the  distance,  and  5  minutes  is  the 
time.  Applying  the  rule,  3,000  H-  5  =  600  ft.  per  min.  Ans. 

CAUTION. — Before  applying  the  above  or  any  of  the  succeeding 
rules,  care  must  be  taken  to  reduce  the  values  given  to  the  denomi- 
nations required  in  the  answer.  Thus,  in  the  above  example,  if  the 
velocity  is  required  in  feet  per  second  instead  of  in  feet  per  minute, 
the  5  minutes  must  be  reduced  to  seconds  before  dividing.  The  oper- 
ation will  then  be:  5  minutes  =  5  X  60  =  300  seconds.  Applying  the 
rule,  3,000  -H  300  =  10  ft.  per  sec.  Ans. 

If  the  velocity  is  required  in  inches  per  second,  it  is  necessary  to 
reduce  the  3,000  feet  to  inches  and  the  5  minutes  to  seconds,  before 
dividing.  Thus,  3,000  feet  X  12  =  36,000  inches.  5  minutes  X  60 
=  300  seconds.  Now  applying  the  rule,  36,000  -f-  300  =  120  in.  per 
Sfec.  Ans. 

EXAMPLE  2. — A  railroad  train  travels  50  miles  in  l£  hours ;  what  is 
its  average  velocity  in  feet  per  second  ? 

SOLUTION. — Reducing  the  miles  to  feet  and  the  hours  to  seconds, 
50  miles  X  5,280  =  264,000  feet.  H  hours  X  60  X  60  =  5,400  seconds. 
Applying  the  rule,  264,000  -*-  5,400  =  48|  ft.  per  sec.  Ans. 

39.  If   the  uniform  velocity  (or  the  average  velocity) 
and  the  time  are  given,  and  it  is  required  to  find  the  distance 
that  a  body  having  the  given  velocity  will  travel  in  the 
given  time : 

Rule  2. — Multiply  the  velocity  by  the  time. 
Or,  d=vt. 

EXAMPLE  1. — The  velocity  of  sound  in  still  air  is  1,092  feet  per 
second;  how  many  miles  will  it  travel  in  16  seconds  ? 

SOLUTION. — Reducing  the  1,092  feet  to  miles',  1,092  H-  5,280  =  \\\\. 
Applying  the  rule,  £§§§  X  16  =  3.31  mi.,  nearly.  Ans. 

EXAMPLE  2. — The  piston  speed  of  an  engine  is  11  feet  per  second, 
how  many  miles  does  the  piston  travel  in  1  hour  and  15  minutes  ? 

SOLUTION. —  1  hour  and  15  minutes  reduced  to  seconds  =  4,500 
seconds  =  the  time.  11  feet  reduced  to  railes  =  ^STF  mile  =  velocity 
in  miles  per  second.  Applying  the  rule,  ^H^  X  4,500  =  9.375  mi.  Ans. 

40.  If   the    distance    through  which    a   body  moves    is 
given,  and  also  its  average  or  uniform  velocity,  and  it  is 
desired  to  know  how  long  it  takes  the  body  to  move  through 
the  given  distance: 


§  4  PRINCIPLES  OF  MECHANICS.  9 

Rule  3. — Divide  the  distance,  or  space  passed  over,  by  the 
velocity. 

'.-£ 

EXAMPLE  1. — Suppose  that  the  radius  of  the  crank  of  a  steam  engine 
is  15  inches  and  that  the  shaft  makes  120  revolutions  per  minute;  how 
long  will  it  take  the  crankpin  to  travel  18,849.6  feet  ? 

SOLUTION. — Since  the  radius,  or  distance  from  the  center  of  the 
shaft  to  the  center  of  the  crankpin,  is  15  inches,  the  diameter  of  the 
circle  it  moves  in  is  15  inches  X  2  =  30  inches  =  2.5  feet.  The  circum- 
ference of  this  circle  is  2.5  X  3.1416  =  7.854  feet.  7.854  X  120  =  942.48 
feet  =  distance  that  the  crankpin  travels  in  1  minute  =  velocity  in  feet 
per  minute.  Applying  the  rule,  18,849.6  -H  942.48  =  20  min.  Ans. 

EXAMPLE  2. — A  point  on  the  rim  of  an  engine  flywheel  travels  at  the 
rate  of  150  feet  per  second ;  how  long  will  it  take  it  to  travel  45,000  feet  ? 
SOLUTION. — Applying  the  rule, 

45,000  -f-  150  =  300  sec.  =  5  min.     Ans. 


EXAMPLES   FOR   PRACTICE. 

1.  A  locomotive  has  drivers  80  inches  in  diameter.     If  they  make 
293  revolutions  per  minute,  what  is  the  velocity  of  the  train  in  (a)  feet- 
per  second  ?  (V)  miles  per  hour  ?  j  («)     102.277  ft.  per  sec. 

IS'1(£)       69. 734  mi.  per  hr. 

2.  Assuming  the  velocity  of  steam  as  it  enters  the  cylinder  to  be 
900  feet  per  second,  how  far  can  it  travel,  if  unobstructed,  during  the 
time  the  flywheel  of  an  engine  revolves  7  times,   if  the  number  of 
revolutions  per  minute  are  120  ?  Ans.  3,150  ft. 

3.  The  average  speed  of  the  piston  of  an  engine  being  528  feet  per 
minute,  how  long  will  it  take  the  piston  to  travel  4  miles  ? 

Ans.  40  min. 

4.  A  speed  of  40  miles  per  hour  equals  how  many  feet  per  second  ? 

Ans.  58f  ft. 

5.  The  earth  turns  around  once  in  24  hours.     If  the  diameter  be 
taken  as  8,000  miles,  what  is  the  velocity  of  a  point  on  the  equator,  in 
miles  per  minute  ?  Ans.  17.45£  mi.  per  min. 

6.  The  stroke   of  an  engine   is  28  inches.     If  the  engine  makes 
11,400  strokes  per   hour,    (a)  what   is  its  speed  in   feet  per  minute  ? 
(£)  how  far  will  this  piston  travel  in  11  minutes  ? 

Ans  -i  (a}    448*  ft'  per  min' 
\  (b)    4,876  ft.  8  in. 


10  PRINCIPLES  OF  MECHANICS.  §  4 

FORCE. 


GENERAL  PRINCIPLES. 

41.  Force  is  known  only  by  the  effects  it  produces  on 
matter.     Forces  cannot,  therefore,  be  compared  in  the  same 
way  that  quantities  of  matter  or  distances  are  compared ; 
the  only  way  in  which  forces  can  be  examined  in  reference 
to  one  another  is  by  noting   their   relative   effects  in  the 
production  of  motion  or  a  change  of  state  in  matter. 

The  most  familiar  conception  of  force  is  that  of  a  push  or 
pull  that  tends  to  produce  or  destroy  motion.  If  the  force 
is  great  enough,  its  effect  is  seen  in  a  change  in  the  state 
of  motion  of  the  body  on  which  it  acts;  that  is,  it  either 
produces  motion  in  the  body  or  destroys  some  or  all  of  the 
motion  already  existing.  If,  however,  the  body  acted  on 
by  a  force  is  so  situated  that  the  force  applied  to  it  is 
opposed  by  a  resisting  force  of  equal  magnitude,  no  change  in 
motion  is  produced.  In  this  case  the  force  is  not  perceiv- 
able, unless  some  other  force  is  introduced  whose  effects 
will  reveal  the  existence  of  the  first  force. 

42.  Forces  are  called  by  various  names  according  to  the 
ways  in  which  they  manifest  themselves.     Manifestations  of 
force  are:   attraction,  repulsion,  coJiesion,  adhesion,  accelera- 
tion, retardation,  resistance,  etc.,   and  the  forces  producing 
these  manifestations  are  called  attractive,  repulsive,  cohesive, 
adhesive,  accelerating,  retarding,  resisting,  etc.  forces. 

43.  Comparison  of  Forces. — In  considering  the  effects 
of  a  force  on  a  body,  some  standard  of  comparison  must  be 
used.     The  standard  most  commonly  adopted  in  English- 
speaking  countries  is  the  pound,  which  is  the  force  required 
to  raise  a  standard  mass  of  matter  from  the  ground  under 
certain  specified  conditions. 

In  practice,  force  is  always  regarded  as  a  pressure;  that 
is,  a  force  is  considered  the  equivalent  of  the  pressure 
exerted  by  a  weight.  For  example,  the  effect  of  a  force  of 


§  4  PRINCIPLES  OF  MECHANICS.  11 

20  pounds  acting  upon  a  body  is  the  same  as  the  pressure 
of  20  pounds  exerted  by  a  weight  of  20  pounds. 

44.  In  order  that  we  may  compare  the  effect  of  a  force 
on  a  body  with  that  of  another  force  on  another  body,  it 
is  necessary  that  the  following  three  conditions  be  fulfilled 
in  regard  to  both  bodies: 

1.  The  point  of  application,  or  point  at  which  the  force 
acts  upon  the  body,  must  be  known. 

2.  The  direction  of  the  force,  or,  what  is  the  same  thing, 
the  straight  line  along  which  the  force  tends  to  move  the 
point  of  application,  must  be  known. 

3.  The  magnitude  or  value  of  the  force  in  comparison 
with  the  given  standard  must  be  known. 

45.  Reduction  of  Forces. — When  a  number  of  forces 
act  upon  a  body  and  produce  a  certain  effect,  it  is  often 
necessary  to  find  a  single  force  that,  when  substituted  for 
the  given  forces,  will  produce  the  same  effect.     In  order  to 
find  the  point  of  application,  the  direction,  and  the  magni- 
tude of  this  single  force,  it  is  necessary  to  know  the  above 
three  conditions  of  every  one  of  the  given  forces. 


NEWTON'S  L.AWS  OF  MOTION. 

46.  The  fundamental  principles  of  the  relations  between 
force  and  motion  were  first  stated  by  Sir  Isaac  Newton,  and 
are  called  Newton  s  Three  Laws  of  Motion.  They  are  as 
follows: 

1.  All  bodies  continue  in  a  state  of  rest,  or  of  uniform 
motion  in  a  straight  /me,  unless  acted  upon  by  some  external 
force  that  compels  a  change. 

2.  Every  motion  or  change  of  motion  is  proportional  to  the 
acting  force,  and  takes  place  in  the  direction  of  the  straight 
line  along  which  the  force  acts. 

3.  To  every  action  there  is  always  opposed  an  equal  reaction. 
These  laws  in  their  accepted  form,  as  just  given,  have 

been  more  or  less  indirectly  derived  from  experiment,  but 


12  PRINCIPLES  OF  MECHANICS.  §  4 

they  are  so  comprehensive  as  to  defy  complete  experimental 
verification. 

47.  Exemplification  of  the  First  Law. — In  the  first  law 
of  motion,  it  is  stated  that  a  body  once  set  in  motion  by  any 
force,  no  matter  how  small,  will  move  forever  in  a  straight 
line,  and  always  with  the  same  velocity,  unless  acted  upon 
by  some  other  force  that  compels  a  change.     It  is  not  pos- 
sible to  actually  verify  this  law,  on  account  of  the  earth's 
attraction  for  all  bodies,  but,  from  astronomical  observa- 
tions, we  are  certain  that  the  law  is  true.     This  law  is  often 
called  the  law  of  inertia. 

48.  The  word  inertia  is  so  abused  that  a  full  under- 
standing of  its  meaning  is  important.     Inertia  is  not  a  force, 
although  it  is  often  so  called.     If  a  force  acts  upon  a  body 
and  puts  it  in  motion,  the  effect  of  the  force  is  stored  in  the 
body,  and  a  second  body,  in  stopping  the  first,  will  receive  a 
blow  equal  in  every  respect  to  the  original  force,  assuming 
that  there  has  been  no  resistance  of  any  kind  to  the  motion 
of  the  first  body. 

It  is  dangerous  for  a  person  to  jump  from  a  fast-moving 
train,  for  the  reason  that,  since  his  body  has  the  same  veloc- 
ity as  the  train,  it  has  the  same  force  stored  in  it  that  would 
cause  a  body  of  the  same  weight  to  take  the  same  velocity 
as  the  train,  and  the  effect  of  a  sudden  stoppage  is  the 
same  as  the  effect  of  a  blow  necessary  to  give  the  person 
that  velocity. 

By  "  bracing  "  himself  and  jumping  in  the  same  direction 
that  the  train  is  moving,  and  running,  he  brings  himself 
gradually  to  rest,  and  thus  reduces  the  danger.  If  a  body 
is  at  rest,  it  must  be  acted  upon  by  a  force  in  order  to  be 
put  in  motion,  and,  no  matter  how  great  the  force  may  be, 
it  cannot  be  instantly  put  in  motion. 

The  resistance  thus  offered  to  being  put  in  motion  is  com- 
monly, but  erroneously,  called  the  resistance  of  inertia.  It 
should  be  called  the  resistance  due  to  inertia. 

49.  Exemplification  of  the  Second  !Law. — From  the 
second  law,  we  see  that  if  two  or  more  forces  act  upon  a 


PRINCIPLES  OF  MECHANICS. 


18 


body,  their  final  effect  on  the  body  will  be  in  proportion 
to  their  magnitudes  and  to  the  directions  in  which  they 
act. 

Thus,  if  the  wind  is  blowing  due  west  with  a  velocity  of 

50  miles  per  hour,  and 
a  ball  is  thrown  due 
north  with  the  same 
velocity,  or  50  miles  per 
hour,  the  wind  will  car- 
ry the  ball  west  while 
the  force  of  the  throw 
is  carrying  it  north,  and 
the  combined  effect  will 
be  to  cause  it  to  move 
northwest. 

The  amount  of  depar- 
ture from  due  north  will 
be  proportional  to  the 
force  of  the  wind  and 
independent  of  the  ve- 
locity due  to  the  force 
of  the  throw. 

5O.  In  Fig.  1  a  ball  e 
is  supported  in  a  cup, 
the  bottom  of  which  is 
attached  to  the  lever  o 
in  such  a  manner  that  o 
will  swing  the  bot- 
tom horizontally  and 
allow  the  ball  to  drop. 
Another  ball  b  rests  in 
a  horizontal  groove  that 
is  provided  with  a  slit 
in  the  bottom.  A  swing- 
ing arm  is  actuated  by  the  spring  d  in  such  a  manner  that, 
when  drawn  back,  as  shown,  and  then  released,  it  will  strike 
the  lever  o  and  the  ball  b  at  the  same  time.  This  gives  b 


14  PRINCIPLES  OF  MECHANICS.  §  4 

an  impulse  in  a  horizontal  direction,  and  swings  o  so  as  to 
allow  e  to  fall. 

On  trying  the  experiment,  it  is  found  that  b  follows  a 
path  shown  by  the  curved  dotted  line,  and  reaches  the  floor 
at  the  same  instant  as  e,  which  drops  vertically.  This 
shows  that  the  force  that  gave  the  first  ball  its  horizontal 
movement  had  no  effect  on  the  vertical  force  that  com- 
pelled both  balls  to  fall  to  the  floor;  the  vertical  force  pro- 
duces the  same  effect  as  if  the  horizontal  force  had  not 
acted.  The  second  law  may  also  be  stated  as  follows:  A 
force  has  the  same  effect  in  producing  motion,  whether  it  acts 
upon  a  body  at  rest  or  in  motion,  and  whether  it  acts  alone  or 
with  other  forces. 

51.  Exemplification  of  the  Third  Law. — The  third 
law  states  that  action  and  reaction  are  equal.  A  man  can- 
not lift  himself  by  his  boot  straps  for  the  reason  that  he 
presses  downwards  with  the  same  force  that  he  pulls 
upwards;  the  downward  reaction  equals  the  upward  action, 
and  is  opposite  to  it. 

In  springing  from  a  boat,  we  must  exercise  caution  or 
the  reaction  will  drive  the  boat  from  the  shore.  When  we 
jump  from  the  ground,  we  tend  to  push  the  earth  from  us, 
while  the  earth  reacts  and  pushes  us  from  it. 

EXAMPLE. — Two  men  pull  on  a  rope  in  opposite  directions,  each 
exerting  a  force  of  100  pounds;  what  is  the  force  that  the  rope 
resists  ? 

SOLUTION. — Imagine  the  rope  to  be  fastened  to  a  tree,  and  that  one 
man  pulls  with  a  force  of  100  pounds.  The  rope  evidently  resists 
100  pounds.  According  to  Newton's  third  law,  the  reaction  of  the  tree 
is  also  100  pounds.  Now,  suppose  the  rope  to  be  slacked,  but  that  one 
end  is  still  fastened  to  the  tree ;  the  second  man  then  takes  hold  of  the 
rope  near  the  tree,  and  pulls  with  a  force  of  100  pounds,  the  first  man 
pulling  as  before.  The  resistance  of  the  rope  is  100  pounds,  as  before, 
since  the  second  man  merely  takes  the  place  of  the  tree.  He  is 
obliged  to  exert  a  force  of  100  pounds  to  keep  the  rope  from  slipping 
through  his  fingers.  If  the  rope  is  passed  around  the  tree,  and  each 
man  pulls  an  end  with  a  force  of  100  pounds  in  the  same  and  parallel 
directions,  the  stress  in  the  rope  is  100  pounds,  as  before,  but  the  tree 
must  resist  the  pull  of  both  men,  or  200  pounds. 


§  4  PRINCIPLES  OF  MECHANICS.  15 

52.  Dynamics,  also  called  kinetics,  is  that  branch  of 
mechanics  that  treats  of  forces  and  their  effects  when  they 
produce  a  change  in  motion  in  the  bodies  on  which  they  act. 

53.  Statics  is  that  branch  of  mechanics  that  treats  of 
forces  and  their  effects  when  they  do  not  produce  a  change 
in  motion  in  the  bodies  on  which  they  act. 


GRAVITATION  AND  WEIGHT. 

54.  Every  body  in  the  universe  exerts  a  certain  attract- 
ive force  on  every  other  body,  which  tends  to  draw  the  two 
together.     To  scientists,  this  attractive  force  is  known  as 
gravitation. 

If  a  body  is  held  in  the  hand,  a  downward  pull  is  felt, 
and  if  the  hold  is  loosened,  the  body  will  fall  to  the  ground. 
This  pull,  which  we  commonly  call  weight,  is  the  attrac- 
tion between  the  earth  and  the  body. 

55.  The  attraction  between  the  earth  and  bodies  at  or 
near  its  surface  is  denoted  by  the  term  force  of  gravity. 
This  attraction  is  generally  considered  as  acting  along  the 
line  joining  the  center  of  gravity  of  the  body  and  the  center 
of  the  earth.     By  center  of  gravity  is  meant  that  point  of 
a  body  at  which  its  whole  weight  may  be  assumed  to  be 
concentrated. 

56.  The  weight  of  a  body  is   directly  proportional  to 
the  force  of  gravity.     From  this  it  follows  that  the  weight 
of  a  body  can  only  be  uniform  everywhere  if  the  force  of 
gravity   is   uniform.     As   a   matter   of   fact,  the   force   of 
gravity    varies    in    different    locations;    consequently,    the 
weight  of  the  body  is  not  the  same  at  all  points  on  the  surface 
of  the  earth.     This  has  been  conclusively  shown  by  sensitive 
spring  balances. 

ACCELERATING  AND  RETARDING  FORCES. 

57.  According  to  the  first  law  of  motion,  if  a  body  is 
set  in  motion  by  a  force  and  the  force  then  ceases  to  act, 
the  body  will  continue  to  move  at  the  rate  it  had  at  the 


16  PRINCIPLES  OF  MECHANICS.  §  4 

instant  the  action  of  the  force  was  discontinued,  unless 
acted  upon  by  some  other  force. 

If,  however,  a  force  acts  upon  a  body  for  a  given  period 
of  time,  say  1  second,  and  imparts  to  it  a  certain  amount  of 
motion,  and  then,  instead  of  ceasing  to  act,  acts  with  the 
same  intensity  during  the  next  second,  it  will  impart  to  the 
body  an  increase  in  velocity  equal  to  the  velocity  imparted 
during  the  first  second.  Then,  the  velocity  at  the  end  of 
the  second  second  will  be  twice  that  at  the  end  of  the  first. 
During  the  third  second  a  like  increase  in  velocity  will  be 
produced,  making  the  velocity  at  the  end  of  the  third  sec- 
ond three  times  as  great  as  at  the  end  of  the  first.  This 
uniform  increase  in  velocity  will  continue  as  long  as  the 
constant  force  continues  to  act  on  the  body.  A  constant 
force  when  producing  a  constant  acceleration  is  called  a 
constant  accelerating  force. 

58.  If  a  constant  force  is  applied  to  a  body  in  motion  in 
such  a  manner  that  it  opposes  the  motion,  its  effect  will  be 
to  reduce  the  motion  by  a  certain  amount,  which  will  be 
the  same  for  each  second  during  which  it   acts.     In  this 
case,  the  force  is  called  a  constant  retarding  force. 

We  thus  see  that  the  effect  of  a  constant  force  acting 
upon  a  body  in  motion  is  to  produce  a  uniform  acceleration 
or  retardation  in  the  velocity  of  motion  of  the  body,  it 
being  assumed  that  the  motion  of  the  body  is  not  opposed 
by  varying  resisting  forces. 

59.  Acceleration  Due  to  the  Force  of  Gravity. — If  a 

body  falls  freely  under  the  action  of  the  force  of  gravity, 
its  velocity  will  increase  at  a  uniform  rate;  in  other  words, 
it  will  be  accelerated.  Since  the  force  of  gravity  varies  in 
different  localities,  it  follows  that  the  acceleration  produced 
by  it  is  not  everywhere  the  same.  The  greatest  range  in 
the  acceleration  due  to  the  force  of  gravity  in  the  United 
States  is  from  a  minimum  of  about  32.089  feet  per  second 
up  to  a  maximum  of  about  32.186  feet  per  second  for  each 
second.  In  the  latitude  of  Scranton,  Pa.,  and  at  the  level 
of  the  sea,  the  acceleration  is  nearly  32.16  feet  per  second; 


§  4  PRINCIPLES  OF  MECHANICS.  17 

this  value  will  be  used  in  all  calculations  in  this  Course  that 
involve  the  use  of  acceleration  due  to  the  force  of  gravity. 
In  accordance  with  the  practice  of  most  scientific  writers, 
we  will  denote  the  acceleration  due  to  the  force  of  gravity 
by  the  letter  g. 

60.  Mass. — If  the  weight  of   a  body  at  any  place,   as 
determined  by  a  spring  balance,  is  divided  by  the  accelera- 
tion due  to  the  force  of  gravity  at  that  place,  a  numerical 
value  will  be  obtained  that,  for  the  same  body,  will  be  the 
same  wherever  it  may  be  weighed.     This  quotient  is  called 
the  mass  of  the  body,  and  is  generally  designated  by  the 
letter  m. 

Rule  4. —  To  find  the  mass  of  a  body,  divide  its  weight  by 
the  acceleration  due  to  the  force  of  gravity. 

Let  W=  the  weight  of  a  body; 

g  =  acceleration  due  to  gravity; 
m  =  mass  of  the  body. 

Then,  m  —  — . 

g 

EXAMPLE. — What  is  the  mass  of  a  body  weighing  96.48  pounds  ? 

q/>  AQ 
SOLUTION. — Applying  the  rule,  m  =  ^TQ  =  3.     Ans. 

61.  Law    of    Gravitation. — The    attractive    force   by 
which  one  body  tends  to  draw  another  body  toward  it  is 
directly  proportional  to  its  mass  and  inversely  proportional 
to  the  square  of  the  distance  between  their  centers  of  gravity. 

62.  Laws  of  Weight. — 

1.  Bodies  weigh  most  at  the  surface  of  the  earth.     Below 
the  surface,  the  weight  decreases  directly  as  the  distance  to 
the  center  of  the  earth  decreases.  * 

2.  Above  the  surface,  the  weight  decreases  inversely  as  the 
square  of  the  distance. 

63.  Change  in  Motion  of  a  Body. — A  change  in  the 
motion  of  a  body  cannot  take  place  without  the  action  of 
an  accelerating  or  retarding  force.     The  force  required  to 


18  PRINCIPLES  OF  MECHANICS.  §  4 

produce  a  given  acceleration  or  retardation  in  a  body  is 
given  by  the  following  rule,  where 

.  f  =  force  in  pounds  ; 
a  =  acceleration  or  retardation  in  feet  per  second. 

Rule  5.  —  Multiply  the  mass  of  the  body  by  the  accelera- 
tion, or  retardation,  in  feet  per  second. 

Or,  f=  ma. 

W  W 

Since  m  =  —  (see  rule  4),  this  may  also  be  written/"  =  —  a. 

EXAMPLE.  —  What  force  will  be  required  to  give  a  body  weighing 
90  pounds  an  acceleration  of  5  feet  per  second  ? 
SOLUTION.—  Applying  rule  5,  we  get 

QA 

5  =  13.99+  lb.,  say  14  lb.     Ans. 


64.  According  to  the  first  law  of  motion,  a  body  in 
motion  not  acted  upon  by  any  external  force  will  continue 
its  motion  without  any  further  application  of  a  force.  In 
practice,  however,  the  motion  of  a  body  is  always  opposed 
by  some  resisting  force  or  forces.  According  to  the  third 
law  of  motion,  the  force  required  to  overcome  the  resistance 
is  equal  to  the  resistance. 

The  opposing  forces  are  usually  constant,  or  nearly  so. 
Taking  the  opposing  forces  into  account,  the  actual  force 
required  to  accelerate  a  body  meeting  with  resistance  will 
be  the  sum  of  the  accelerating  force  and  the  opposing  forces. 

ILLUSTRATION.  —  Imagine  a  weight  of  321.6  pounds  to  be  lying  on  a 
smooth  plane  surface.  Assume  that  it  has  been  determined  experi- 
mentally that  a  force  of  100  pounds  is  required  to  be  exerted  con- 
tinually to  overcome  the  friction  between  the  weight  and  the  surface. 
What  force  will  be  required  to  produce  an  acceleration  of  2  feet  per 
second  ? 

By  rule  5,  the  accelerating  force  is    ^-f^  X  2  =  20  pounds.      As  a 


force  of  100  pounds  must  be  exerted  continually  to  overcome  the  resist- 
ance due  to  friction,  a  force  of  100  +  20  =  120  pounds  will  be  required 
to  produce  the  required  acceleration. 

65.     The  question  is  often  asked,  what  force  is  required 
to  start  a  flywheel  and  keep  it  going  at  a  stated  number  of 


§  4  PRINCIPLES  OF  MECHANICS.  19 

revolutions  per  minute  ?  This  question,  or  similar  questions, 
cannot  be  answered  without  knowing  the  time  in  which  the 
flywheel  is  to  attain  the  given  speed.  An  accelerating 
force  depending  on  the  mass  of  the  wheel,  its  diameter,  the 
distribution  of  the  material  of  which  it  is  made,  the  time  of 
acceleration,  and  the  constant  resistances,  is  required  to 
bring  the  wheel  up  to  its  speed.  When  this  speed  has  been 
attained,  the  force  required  to  keep  it  going  will  be  that 
required  to  overcome  the  frictional  and  air  resistances. 


MOMENTUM. 

66.  Experience  teaches  us  that   the  same  force  acting 
upon  bodies  of  different  weights  produces  different  effects. 
For  example,  if  a  given  force  imparts  a  velocity  of  10  feet 
per  second  in  a  certain  time  to  a  body  weighing  1  pound,  we 
know  from  experience  and  observation  that  it  cannot  impart 
the  same  velocity  in  the  same  time  when  acting  upon  a  body 
weighing  1,000  pounds. 

Scientists  have  shown  that  the  velocity  imparted  to  a  body 
in  a  given  time  by  a  force  varies  directly  as  the  force  and 
inversely  as  the  mass  of  the  body.  Hence,  forces  may  be 
compared  with  one  another  by  comparing  their  effects  in 
imparting  velocities  to  bodies  whose  masses  are  known. 

67.  The  product  obtained  by  multiplying  the  mass  of  a 
body  by  its  velocity  in  feet  per  second  is  called  the  momen- 
tum of  the  body ;  it  represents  the  magnitude  of  the  force 
that  will  produce  the  given  velocity  in  the  body  in  1  second. 
Hence,  we  may  call  momentum  the  time  effect  of  a  force. 

68.  According  to  the  third  law  of  motion,  action  and 
reaction    are   equal   to   each    other.     Consequently,    if   the 
force  required  to  produce  a  stated  momentum  in  a  given 
time  is  known,  it  is  likewise  known  what  force  is  required  to 
destroy  this  momentum,  or  to  bring  the  body  to  rest,  in  an 
equal  period  of  time. 

When  a  liquid  body  is  flowing  in  a  stream,  as  from  a 
nozzle,  the  weight  to  be  considered  in  problems  involving 


20  PRINCIPLES  OF  MECHANICS.  §  4 

momentum  is  the  weight  of  the  liquid  discharged  in  1  second. 
For  example,  let  it  be  required  to  estimate  the  force  with 
which  a  man  must  hold  the  nozzle  of  a  fire  hose  to  prevent 
its  slipping  through  his  hands  when  a  stream  of  water  issues 
from  it  with  a  velocity  of  20  feet  per  second,  the  area  of  the 
opening  in  the  nozzle  being  1  square  inch  and  the  weight  of 
a  cubic  inch  of  water  .0361  pound.  The  volume  of  water 
discharged  in  1  second  is  1  X  12  X  20  =  240  cubic  inches, 
and  since  the  weight  of  1  cubic  inch  is  .0361  pound,  the 
weight  discharged  per  second  is  240  X  .0361  =  8.664  pounds. 

The  momentum  of  the  stream  is  -^  ^—  X  20  =  5.38  pounds. 

o*.lo  , 

This  is  the  constant  force  required  to  give  a  body  of  water 
weighing  8.664  pounds  a  velocity  of  20  feet  per  second,  in 
1  second ;  it  also  represents  the  magnitude  of  the  reaction 
on  the  nozzle,  and  the  man  must  hold  the  nozzle  with  a  force 
equal  to  the  reaction,  or  5.38  pounds,  in  order  to  prevent  its 
slipping  through  his  hands. 


WORK,  POWER,  AOT> 


WORK. 

69.  Work  is  the  overcoming  of  resistance  continually 
occurring  along  the  path  of  motion. 

Motion  in  itself  is  not  work ;  a  force  must  overcome  a 
resistance  in  order  that  work  may  be  done. 

70.  Unit  of  Work. — The  unit  by  which  the  work  done 
by  a  force  is  measured  is  the  work  done  in  overcoming  a 
resistance  of  1  pound  through  a  space  of  1  foot;  this  unit  is 
called  a  foot-pound.     According  to  the  definition,  it  may 
be  considered  as  the   force  required  to  raise  1  pound  1  foot 
vertically.     All  work  is  measured  by  this  standard. 

A  horse  going  up  a  hill  does  an  amount  of  work  equal  to 
its  own  weight,  plus  the  weight  of  the  wagon  and  its  con- 
tents, plus  the  frictional  resistances  reduced  to  an  equivalent 


§  4  PRINCIPLES  OF  MECHANICS.  21 

weight,  multiplied  by  the  vertical  height  of  the  hill.  Thus, 
if  the  horse  weighs  1,200  pounds,  the  wagon  and  contents 
1,200  pounds,  and  the  frictional  resistances  equal  400  pounds, 
then  if  the  vertical  height  of  the  hill  is  100  feet,  the  work 
done  is  equal  to  (1,200  +  1,200  +  400)  X  100  =  280,000  foot- 
pounds. 

Rule  6. — In  all  cases  the  force  (or  resistance]  multiplied 
by  the  distance  through  wJncJi  it  acts  equals  the  work.  If  a 
weight  is  raised,  the  weight  multiplied  by  the  vertical  height 
of  the  lift  equals  the  work. 

71.  The  total  amount  of  work  done  in  overcoming  a 
given  resistance  through  a  given  distance  is  independent  of 
time;  that  is,  it. is  immaterial  whether  it  takes  1  minute  or 
1  year  in  which  to  do  it ;  but  in  order  to  compare  the  rate 
at  which  work  is  done  by  different  machines  with  a  common 
standard,  time  must  be  considered.  If  one  machine  does  a 
certain  amount  of  work  in  10  minutes  and  another  machine 
does  exactly  the  same  amount  of  work  in  5  minutes,  the 
second  machine  can  do  twice  as  much  work  as  the  first  in 
an  equal  period  of  time. 


POWER. 

72.  Power  is  a  term  used  to  denote  the  rate  at  which 
work  is  done. 

73.  The  common  unit  used  for  expressing  the  rate  at 
which  work  is  done  is  the  horsepower. 

One  horsepower  is  33,000  foot-pounds  of  work  per  minute; 
in  other  words,  it  is  33,000  pounds  raised  vertically  1  foot  in 
1  minute,  or  1  pound  raised  vertically  33,000  feet  in  1  min- 
ute, or  any  combination  that  will,  when  multiplied  together, 
give  33,000  foot-pounds  in  1  minute.  Thus,  the  work  done 
in  raising  110  pounds  vertically  5  feet  in  1  second  is  a 
horsepower;  for,  since  in  1  minute  there  are  60  seconds, 
110  X  5  X  60  =  33,000  foot-pounds  in  1  minute. 

EXAMPLE.— -In  a  steam  engine  the  force  impelling  the  piston  for- 
wards and  backwards  is  10,000  pounds.  This  force  overcomes  the 

H.  8.    /.— 12 


22  PRINCIPLES  OF  MECHANICS.  §  4 

resistance  due  to  the  load  at  the  rate  of  600  feet  per  minute ;  that  is,  it 
moves  the  piston  back  and  forth  at  that  rate.  What  is  the  horsepower 
of  the  engine  ? 

SOLUTION.— According  to  rule  6,  the  work  done  is  10,000  X  600 
=  6,000,000  foot-pounds  per  minute.  Then,  as  33,000  foot-pounds  per 
minute  represent  a  horsepower,  the  horsepower  of  the  engine  is 
6,000,000  H-  33,000  =  181.818.  Ans. 


ENERGY. 

74.  Energy  is  a  term  used  to  express  the  ability  of  an 
agent  to  do  work. 

75.  Kinetic  Energy. — If  we  have  a  body  at  rest,  a  cer- 
tain amount  of  force  must  be  exerted  and  a  certain  amount 
of  work  must  be  done  to  set  it  in  motion.     A  part  of  this 
force  will  be  required  to  overcome  those  resistances  outside 
of  the  body,  such  as  friction  and  the  resistance  of  the  air, 
that  oppose  the  motion  of  all  bodies  with  which  we  have  to 
do ;  another  part  acts  to  overcome  the  inertia  of  the  body, 
to  start  it  from  its  state  of  rest,  and  give  it  motion  (see 
Newton's  first  law).     The  force  that  overcomes  the  resist- 
ance due  to  the  inertia  of  the  body  does  work,  and  the  work 
so  performed  is  stored  in  the  body;  in  being  brought  to 
rest,   the  body  is  capable  of  overcoming  a  resistance   and 
of  doing  an   amount  of  work  exactly  equal   to  the  work 
expended  in  giving  it  motion.     The  ability  that  the  moving 
body  has  to  do  work  while  being  brought  to  rest   is  called 
the  kinetic  energy  of  the  body. 

76.  Rule   for    the    Energy   of  a   Moving   Body. — 

Let  w—  weight  of  body  in  pounds; 

v  =  its  velocity  in  feet  per  second; 
E  =  kinetic  energy  in  foot-pounds. 

Then,  the  kinetic  energy  of  a  moving  body  may  be  found 
as  follows: 

Rule  7. — Multiply  the  weight  of  the  body  by  the  square 
of  its  velocity  and  divide  the  product  by  twice  the  acceleration 
due  to  the  force  of  gravity. 


§  4  PRINCIPLES  OF  MECHANICS.  23 

wv* 


Or,  E  = 


64.32' 


Thus,  if  a  weight  is  raised  a  certain  height,  an  amount, 
of  work  is  done  equal  to  the  product  of  the  weight  and  the 
vertical  height.  If  a  weight  is  suspended  at  a  certain 
height  and  allowed  to  fall,  it  will  do  the  same  amount  of 
work  in  foot-pounds  that  was  required  to  raise  the  weight  to 
the  height  through  which  it  fell. 

EXAMPLE  1.—  If  a  body  weighing  25  pounds  falls  from  a  height  of 
100  feet,  how  much  work  can  it  do  ? 

SOLUTION.  —  Work  =  w  h  =  25  x  100  =  2,500  ft.-lb.     Ans. 

It  requires  the  same  amount  of  work  or  energy  to  stop  a 
body  in  motion  within  a  certain  time  as  it  does  to  give  it 
that  velocity  in  the  same  length  of  time. 

EXAMPLE  2.—  A  body  weighing  50  pounds  has  a  velocity  of  100  feet 
per  second;  what  is  its  kinetic  energy  ? 

SOLUTION.—  Kinetic  energy  =  -^  =  ™£^  =  7.773.63  ft.-lb. 
o4.o^  O4.OA  . 

Ans. 

EXAMPLE  3.—  In  the  last  example,  how  many  horsepower  will  be 

required  to  give  the  body  this  amount  of  kinetic  energy  in  3  seconds  ? 

SOLUTION.  —    1  horsepower  =  33,000  pounds  raised  1  foot  in  1  minute. 

If  7,773.63  foot-pounds  of  work  are  done  in  3  seconds,  in  1  second 

there  will  be  done  —  —  ^  —  =  2,591.21  foot-pounds  of  work.     1  horse- 

power =  33,000  foot-pounds  per  minute  =  33,000  -f-  60  =  550  foot-pounds 
per  second. 

The  number  of  horsepower  required  will  be 

=  4.7113  H.  P.     An, 


77.  Potential  energy  is  latent  energy;  it  is  the  energy 
that  a  body  at  rest  is  capable  of  giving  out  under  certain 
conditions. 

If  a  stone  is  suspended  by  a  string  from  a  high  tower,  it 
has  potential  energy.  If  the  string  is  cut,  the  stone  will  fall 
to  the  ground,  and  during  its  fall  its  potential  energy  will 
change  into  kinetic  energy,  so  that  at  the  instant  it  strikes 
the  ground  its  potential  energy  is  wholly  changed  into  kinetic 
energy 


24  PRINCIPLES  OF  MECHANICS.  §  4 

At  a  point  equal  to  one-half  the  height  of  the  fall,  the 
potential  and  kinetic  energies  are  equal.  At  the  end  of 
the  first  quarter  the  potential  energy  is  three-fourths  and 
the  kinetic  energy  one-fourth;  at  the  end  of  the  third 
quarter  the  potential  energy  is  one-fourth  and  the  kinetic 
energy  three-fourths. 

A  pound  of  coal  has  a  certain  amount  of  potential  energy. 
When  the  coal  is  burned,  the  potential  energy  is  liberated 
and  changed  into  kinetic  energy  in  the  form  of  heat.  The 
kinetic  energy  of  the  heat  changes  water  into  steam,  which 
thus  has  a  certain  amount  of  potential  energy.  The  steam 
acting  on  the  piston  of  an  engine  causes  it  to  move  through 
a  certain  space,  thus  overcoming  a  resistance,  changing  the 
potential  energy  of  the  steam  into  kinetic  energy,  and  thus 
doing  work. 

Potential  energy,  then,  is  the  energy  stored  within  a  body 
that  may  be  liberated  and  produce  motion,  thus  generating 
kinetic  energy  and  enabling  work  to  be  done. 

78.  The  principle  of  conservation  of  energy  teaches 
that    energy,   like    matter,   can  never  be  destroyed.      If    a 
clock  is  put  in  motion,  the  potential  energy  of  the  spring  is 
changed  into  kinetic   energy  of    motion,    which  turns   the 
wheels,  thus  producing  friction. 

The  friction  produces  heat,  which  dissipates  into  the  sur- 
rounding air,  but  still  the  energy  is  not  destroyed — it  merely 
exists  in  another  form. 

79.  Work   of  Acceleration  and   Retardation.  —  The 

theoretical  amount  of  work  that  must  be  done  in  order  to 
start  a  body  from  a  state  of  rest  and  accelerate  it  until  it 
reaches  a  given  velocity  is  equal  to  the  kinetic  energy  of 
the  body  at  the  given  velocity.  Likewise,  the  theoretical 
amount  of  work  that  must  be  done  on  a  moving  body  to 
retard  it  and  finally  bring  it  to  rest  is  equal  to  the  kinetic 
energy  the  moving  body  possessed  at  the  moment  retarda- 
tion began.  The  work  that  must  be  done  in  changing  the 
velocity  of  a  body  is  equal  to  the  difference  in  the  kinetic 
energies  at  the  initial  and  final  velocities.  Since  the  motion 


§  4  PRINCIPLES  OF  MECHANICS.  25 

of  all  bodies  is  opposed  by  some  resisting  force  or  forces,  the 
actual  amount  of  work  required  to  give  a  body  the  given 
velocity  will  be  the  sum  of  the  work  of  acceleration  and  the 
work  required  to  overcome  the  outside  resisting  forces. 

EXAMPLE  1.  —  A  body  weighing  1,000  pounds  is  started  from  rest 
and  is  to  attain  a  velocity  of  88  feet  per  second  in  2  minutes,  passing 
over  a  distance  of  5,280  feet  in  that  time.  If  a  constant  force  of 
120  pounds  must  be  exerted  to  overcome  the  frictional  resistances, 
what  work  must  be  done  ? 

SOLUTION.  —  According  to  this  article,  the  work  required  to  accelerate 

1  000  v  88'^ 
the  body  is        ^  =  120,398  foot-pounds.     As  a  constant  force  of 


120  pounds  must  act  through  a  distance  of  5,280  feet  to  overcome  the 
frictional  resistances,  the  work  done  in  overcoming  friction  is  5,280 
X  120  =  633,600  foot-pounds.  Then,  the  total  amount  of  work  done  is 
120,398  +  633,600  =  753,998  ft.-lb.  Ans. 

EXAMPLE  2.  —  What  horsepower  will  be  required  to  do  the  work  cal- 
culated in  the  last  example  ? 

SOLUTION.  —  As  the  work  is  done  in  2  minutes,  the  horsepower  is 
753,998 


2  X  33,000 


=  11.424  H.  P.,  nearly.     Ans. 


FORCE  OF  A  BL.OW. 

8O.  The  questions  are  frequently  asked,  with  what  force 
will  a  falling  hammer  strike,  or  with  what  force  will  a  pro- 
jectile fired  from  a  gun  strike  an  object  ?  These  questions 
catinot  be  answered  directly,  as  they  are  based  on  a  miscon- 
ception. A  moving  body  possesses  kinetic  energy,  or  ability 
to  do  work,  which  ability  can  only  be  expressed  in  foot- 
pounds, but  not  in  pounds  of  force,  since  the  work  done  by 
the  hammer  or  projectile  in  coming  to  rest  is  not  a  manifes- 
tation of  force,  but  of  energy. 

Work  is  the  product  of  force  into  distance ;  hence,  if  the 
amount  of  work  a  body  has  done  or  is  capable  of  doing  'is 
known,  the  force  can  be  determined  for  each  case  if,  by  some 
means,  it  is  possible  to  determine  exactly  the  distance  in 
which  the  work  is  done.  This  distance  depends  on  vari- 
ous resistances,  such  as  that  due  to  moving  the  object 
struck,  the  resistance  to  penetration,  friction,  the  resistance 


26  PRINCIPLES  OF  MECHANICS.  §4 

to  shearing  or  deformation  of  the  body,  etc.  The  distance 
through  which  these  resisting  forces  act  is  generally  inde- 
terminate, and  since  the  average  of  the  resisting  forces 
varies  generally  with  the  distance,  this  average  resisting 
force  is  also  indeterminate;  hence,  the  force  that,  acting 
through  a  distance,  will  absorb  all  the  kinetic  energy  of  the 
hammer  or  projectile  cannot  be  determined  for  the  practical 
reasons  given. 

COMPOSITION  A1STD  RESOLUTION  OF  FORCES. 

81.  According  to  Art.  44,  in  order  that  forces  may  be 
compared  with  one  another,  three  conditions  must  be  ful- 
filled.    These  conditions  may  all  be  repre- 
sented  by  a  line ;  hence,  we  may  represent 

FIG- 2-  forces  by  lines.     Thus,  in  Fig.  2,  let  A  be 

the  point  of  application  of  the  force,  let  the  length  of  the 
line  A  B  represent  its  magnitude,  and  let  the  arrowhead 
indicate  the  direction  in  which  the  force  acts;  then,  the 
line  A  B  fulfils  the  three  conditions  and  the  force  is  fully 
represented. 

COMPOSITION  OF  FORCES. 

82.  When  two  or  more  forces  act  upon  a  body  at  the 
same  time  along  lines  that  meet  in  a  common  point,  their 
combined  effect  on  the  body  may  be  obtained  by  an  applica- 
tion of  the  principle  of  the  triangle  of  forces. 

In  Fig.  3  (a),  let  A  and  B  be  two  forces  having  the  mag- 
nitudes and  directions  represented  by  the  two  lines.     To  find 

..    A 


the  effect  due  to  the  combined  action  of  these  two  forces, 
draw  in  any  convenient  location  a  line  parallel  to  either  of  the 


§  4  PRINCIPLES  OF  MECHANICS.  27 

lines  representing  the  two  forces,  making  it  equal  in  length, 
to  some  scale,  to  the  magnitude  of  the  force.  Mark  upon 
it  an  arrowhead  pointing  in  the  same  direction  as  the  arrow- 
head on  the  line  representing  the  force.  Then,  from  one 
extremity  of  the  line  just  drawn  draw  a  line  parallel  to  the 
line  representing  the  second  force  and  equal  in  length,  to  the 
same  scale,  to  the  magnitude  of  the  second  force,  and  mark 
the  arrowhead  upon  it.  It  is  essential  that  the  second  line 
be  so  drawn  that  when  passing  over  the  two  lines  with  a 
pencil,  commencing  at  the  beginning  of  either  force,  the 
arrowheads  will  both  point  in  the  same  direction  in  reference 
to  the  direction  of  motion  of  the  pencil. 

The  second  line  may  be  drawn  from  either  end  of  the  first 
line,  but  its  direction  must  be  made  to  fulfil  the  above  abso- 
lutely essential  condition.  Thus,  in  Fig.  3  (6),  the  line  B' 
has  been  drawn  from  the  right-hand  extremity  of  A'.  Start- 
ing at  a  with  a  pencil  and  moving  toward  b,  the  pencil  will 
move  in  the  direction  in  which  the  arrowhead  on  A'  points. 
Passing  over  B'  from  b  to  c,  the  pencil  will  move  in  the 
direction  in  which  the  arrowhead  on  B'  points;  we  thus  see 
that  the  lines  are  drawn  correctly  in  reference  to  each  other. 

In  Fig.  3  (c),  the  line  H'has  been  drawn  from  the  left-hand 
extremity  of  A'.  Starting  at  c  and  following  up  the  lines, 
it  is  seen  that  in  this  case  the  arrowheads  both  point  in  the 
same  direction  relative  to  the  direction  of  motion  of  the 
pencil,  thus  showing  the  lines  to  be  located  correctly. 

The  two  lines  having  been  drawn,  complete  the  triangle 
by  drawing  the  line  C.  This  line,  called  the  resultant, 
represents  the  combined  effect  of  the  two  forces;  it  gives 
the  direction  along  which  the  two  forces  will  act  when  com- 
bined. The  magnitude  of  their  combined  effect  is  found  by 
measuring  this  line  by  the  scale  with  which  A'  and  B'  were 
laid  off.  The  resultant  will  always  have  a  direction  opposite 
in  sense  to  that  of  the  forces;  that  is,  if  we  pass  a  pencil 
around  the  triangle  in  the  direction  in  which  the  arrowheads 
on  the  lines  A'  and  B'  point,  the  arrowhead  on  C,  represent- 
ing the  direction  of  action  of  the  resultant,  must  point  in  a 
direction  opposite  to  that  in  which  the  pencil  moves. 


28  PRINCIPLES  OP  MECHANICS.  §  4 

83.  In  practice  it  is  often  desired  to  find  not  only  the 
magnitude  and  direction,  but  also  the  actual  location  of  the 
resultant,  the  magnitudes  and  lines  of  action  of  the  two 
forces  being  known.  (By  location  of  the  forces  and  result- 
ant is  meant  the  location  of  the  lines  along  which  the  forces 
actually  act.)  This  can  readily  be  done  by  producing  the 
lines  giving  the  location  of  the  forces  until  they  meet  and 
then  drawing  the  triangle  of  forces  with  their  point  of 
intersection  as  the  starting  point. 

ILLUSTRATION. — In  Fig.  4  is  shown  a  head-frame  erected  at  the 
mouth  of  a  deep,  vertical  shaft.  The  hoisting  rope  that  leads  to  the 


Scale  1=1000  Jfc 


FIG.  4. 

hoisting  engine  passes  over  a  sheave  at  the  top  of  the  head-frame.  A 
weight  of  1,500  pounds  hangs  at  the  end  of  the  rope.  Neglecting  the 
weight  of  the  rope  and  sheave,  what  is  the  total  pressure  on  the  bear- 
ings of  the  sheave  ? 

According  to  Art.  51,  the  stress  in  both  parts  of  the  hoisting  rope 
is  1,500  pounds.     The  part   A  B  of  the   rope  supports  the  weight; 


§  4  PRINCIPLES  OF  MECHANICS.  29 

consequently,  the  force  acting  along  A  B  is  downwards.  The  force 
acting  along  CD  is  the  pull  exerted  by  the  engine  and  is  toward  the 
engine.  To  find  the  resultant  of  these  two  forces,  draw  the  triangle  of 
forces.  Choosing  a  scale  of  1  inch  per  1,000  pounds,  draw  the  line  E  F 
parallel  to  CD,  making  it  |{j{$  =  1.5  inches  long.  From  F  draw  FG 
parallel  to  A  B  and  j£gg  =  1.5  inches  long.  Join  E  and  G;  then  EG 
will  be  the  resultant  whose  magnitude  is  measured  by  the  same  scale. 
Measuring  EG,  it  is  found  to  be  2.75  inches  long.  As  each  inch  repre- 
sents 1,000  pounds,  the  magnitude  of  the  resultant  is  2.75  X  1,000 
=  2,750  pounds. 

To  find  the  actual  location  of  the  resultant  in  reference  to  A  B 
and  CD,  produce  both  lines,  as  shown  in  dotted  lines,  until  they 
intersect  at  E'.  Starting  the  triangle  of  forces  at  E',  lay  off  E'  F' 
— 1.5  inches.  From  F'  draw  F'  G'  —  1.5  inches  and  parallel  to  A  B. 
Join  E'  G'.  Upon  measurement  it  is  found  to  be  2,750  pounds.  As 
inspection  shows  that  the  resultant  passes  through  the  center  of  the 
bearings,  the  total  pressure  on  the  bearings  is  2,750  pounds.  Ans. 


RESOLUTION  OF  FORCES. 

84.  Since  two  forces  can  be  combined  to  form  a  single 
resultant  force,  we  may  also  treat  a  single  force  as  if  it  were 
the  resultant  of  two  forces 
whose  action  upon  a  body 
will  be  the  same  as  that  of 
a  single  force.  Thus,  in 
Fig.  5,  the  force  OA  may 
be  resolved  into  two  forces 
O£and£A.  If  the  force 
O  A  acts  upon  a  body 
moving  or  at  rest  upon  a  FIG-  5- 

horizontal  plane,  and  the  resolved  force  O  B  is  vertical 
and  B  A  horizontal,  OB,  measured  to  the  same  scale  as  O  A, 
is  the  magnitude  of  that  part  of  O  A  that  pushes  the  body 
downwards,  while  B  A  is  the  magnitude  of  that  part  of  the 
force  O  A  that  is  exerted  in  pushing  the  body  in  a  horizontal 
direction.  O  B  and  B  A  are  called  the  components  of  the 
force  O  A,  and  when  these  components  are  vertical  and 
horizontal,  as  in  the  present  case,  they  are  called  the  verti- 
cal component  and  the  horizontal  component  of  the  force  O  A. 


30  PRINCIPLES  OP  MECHANICS.  §4 

85.  It  frequently  happens  that  the  position,  magnitude, 
and  direction  of  a  certain  force  is  known,  and  that  it  is 
desired  to  know  the  effect  of  the  force  in  some  direction 
other  than  that  in  which  it  acts.     Thus,  in  Fig.  5,  suppose 
that  O  A  represents,  to  some  scale,  the  magnitude,  direction, 
and  line  of  action  of  a  force  acting  upon  a  body  at  A,  and 
that  it  is  desired  to  know  what  effect  O  A  produces  in  the 
direction  B  A.     Now  B  A,  instead  of  being  horizontal,  as  in 
the  figure,  may  have  any  direction.     To  find  the  value  of 
the  component  of  O  A   which  acts  in  the  direction  B  A,  we 
use  the  following  rule: 

Rule  8. — From  one  extremity  of  the  line  representing  the 
given  force,  draw  a  line  parallel  to  the  direction  in  which 
it  is  desired  that  the  component  shall  act  /  from  the  other 
extremity  of  the  given  force,  draw  a  line  perpendicular  to  the 
component  first  drawn,  and  intersecting  it.  The  length  of 
the  component,  measured  from  the  point  of  intersection  to  the 
intersection  of  the  component  with  the  given  force,  will  be 
the  magnitude  of  the  effect  produced  by  the  given  force  in  the 
required  direction. 

Thus,  suppose  O  A,  Fig.  5,  represents  a  force  acting 
upon  a  body  resting  upon  a  horizontal  plane,  and  it  is 
desired  to  know  what  vertical  pressure  O  A  produces  on  the 
body.  Here  the  desired  direction  is  vertical;  hence,  from 
one  extremity,  as  O,  draw  OB  parallel  to  the  desired  direc- 
tion (vertical  in  this  case),  and  from  the  other  extremity 
draw  A  B  perpendicular  to  OB  and  intersecting  OB  at  B. 
Then  OB,  when  measured  to  the  same  scale  as  O  A,  will  be 
the  value  of  the  vertical  pressure  produced  by  O  A. 

86.  Tangential   Pressure. — One  of  the  most  familiar 
applications   of   the    principle   of   the   resolution   of   forces 
occurring  in  steam  engineering  is  the  case  of  the  connecting- 
rod  and  crank.     When  the  piston  is  at  the  end  of  its  stroke 
and  the  crankpin  is  in  a  line  drawn  through  the  center  of 
the  cylinder  and  crank-shaft,  a  position  that  is  expressed  by 
saying  the  engine  is  "on  the  center,"  the  pressure  of  the 
connecting-rod  on   the  crankpin  acts  directly  against  the 


PRINCIPLES  OF  MECHANICS. 


31 


bearings  of  the  shaft  and  there  is  no  turning  effect  on 
the  pin.  After  the  pin  leaves  the  center,  the  pressure 
exerted  on  it  by  the  connecting-rod  may  be  resolved  into  two 
components:  One  of  these  components  acts  in  the  direction 
of  the  center  line  PO  of  the  crank  (see  Fig.  6)  and  merely 


Scale  1*200O  lb. 


exerts  a  pressure  on  the  bearings  of  the  shaft;  the  other 
component  acts  along  the  line  A  B  at  right  angles  to  the 
center  line  PO  and  tangent  to  the  circle  described  by  the 
crankpin.  This  is  the  force  that  tends  to  turn  the  crank 
and  is  called  the  tangential  pressure  on  the  crankpin. 
When  the  crank  is  on  the  center,  there  is  no  tangential 
pressure  on  the  pin  and  no  tendency  to  turn  the  crank. 
The  tangential  pressure  gradually  increases  as  the  pin  leaves 
the  center  and  becomes  greatest  at  the  point  where  the  con- 
necting-rod and  crank  are  at  right  angles  to  each  other;  it 
then  decreases  until  the  next  center  is  reached.  At 'the 
position  where  the  connecting-rod  and  crank  are  at  right 
angles  to  each  other,  the  tangential  pressure  on  the  pin  is 
equal  to  the  total  pressure  exerted  on  it  by  the  connecting- 
rod,  and  there  is  no  component  in  the  direction  of  the  center 
line  PO  of  the  crank. 


32  PRINCIPLES  OF  MECHANICS.  §  4 

EXAMPLE. — If  a  force  of  3,000  pounds  acts  along  the  connecting-rod  a 
in  the  direction  of  the  arrow  (see  Fig.  6),  what  is  the  tangential  pres- 
sure on  the  crankpin  ? 

SOLUTION. — As  the  tangential  pressure  is  the  pressure  perpendicu- 
lar 'to  the  crank,  draw  A  B  through  the  crankpin  center  at  right 
angles  to  the  center  line  of  the  crank ;  A  B  then  represents  the  line 
along  which  the  tangential  pressure  acts.  Then,  in  any  convenient 
location,  draw  CD  parallel  to  the  connecting-rod.  Choosing  a  scale  of 
2,000  pounds  =  1  inch,  make  the  line  CD  3,000  -H  2,000  =  1|  inches 
long.  From  D  draw  an  indefinite  line  D  E  parallel  to  A  B,  and 
draw  C  E  perpendicular  to  D  E.  Now  ED,  measured  to  the  scale 
adopted,  will  be  the  magnitude  of  the  tangential  pressure  for  the  posi- 
tion of  the  crank  and  connecting-rod  shown  in  the  figure.  Upon 
measurement,  it  is  found  to  be  1.3  inches  long.  Then  1.3  X  2,000 
=  2,600  lb.,  the  tangential  pressure.  Ans. 

87.  When  the  total  pressure  on  the  piston  rod  of  a  steam 
engine  is  known,  the  force  acting  along  the  connecting-rod 
and  the  force  acting  upon  the  guides  can  be  determined  by 
the  following  application  of  the  principle  of  the  resolution 
of  force: 

Draw  a  line,  as  A  B,  Fig.  7,  parallel  to  the  line  of  motion 
of  the  crosshead  a.  Make  its  length,  to  some  scale,  equal  in 


magnitude  to  the  force  impelling  the  piston.  From  one 
extremity  of  A  B  draw  a  line  B  C  parallel  to  the  center  line 
of  the  connecting-rod  b.  From  the  other  extremity  of  A  B 
draw  a  line  at  a  right  angle  to  A  B,  producing  it  until  it 
intersects  the  line  B  C  at  D.  Then,  B  D  represents  the  force 
acting  along  the  connecting-rod  and  A  D  represents  the 
force  acting  upon  the  guides. 


§4  PRINCIPLES  OF  MECHANICS.  33 

FRICTION. 

88.  Friction  is  the  resistance  that   a  body  meets  with 
from  the  surface  upon  which  it  moves. 

89.  The  ratio  between  the  resistance  to  the  motion  of  a 
body  due  to  friction  and  the  perpendicular  pressure  between 
the  surfaces  is  called  the  coefficient  of  friction. 

If  a  weight  W,  as  in  Fig.  8,  rests  upon  a  horizontal  plane, 
and  has  a  cord  fastened  to  it  passing  over  a  pulley  «,  from 


100  Ib. 
W 


10  Ib. 


which  a  weight  Pis  suspended,  then,  if  P  is  just   sufficient 

p 
to  start  W,  the  ratio  of  P  to  W,  or  yr>,  is  the  coefficient  of 

friction  between  Wand  the  surface  it  slides  upon.  The 
weight  W  is  the  perpendicular  pressure,  and  P  is  the  force 
necessary  to  overcome  the  resistance  to  the  motion  of  W 
due  to  friction.  If  W  =  WO  pounds  and  P=  10  pounds,  the 

p 
coefficient  of  friction  for  this  particular  case  would  be  - 


- 
100 

9O.     Laws  of  Friction.  — 

1.  Friction   is  directly  proportional  to   the  perpendicular 
pressure  between  the  two  surfaces  in  contact. 

2.  Friction  is  independent  of  the  extent  of  the  surfaces  in 
contact  when  the  total  perpendicular  pressure  remains  the  same. 

3.  Friction  increases  with  the  roughness  of  the  surfaces. 

4.  Friction  is  greater  between  surfaces  of  the  same  mate- 
rial than  between  those  of  different  materials. 


34  PRINCIPLES  OF  MECHANICS.  §  4 

5.  Friction  is  greatest  at  the  beginning  of  motion. 

6.  Friction  is  greater  between  soft  bodies  than   between 
hard  ones. 

7.  Rolling  friction  is  less  than  sliding  friction. 

8.  Friction  is  diminished  by  polishing'  or  hibricating  the 
surfaces. 

91.  Law  1  shows  why  the  friction  is  so  much  greater  on 
journals  after  they  begin  to  heat  than  before.     The  heat 
causes  the  journal  to  expand,  thus  increasing  the  pressure 
between  the  journal    and   its   bearing,   and,    consequently, 
increasing  the  friction. 

Law  2  states  that,  no  matter  how  small  may  be  the  sur- 
face that  presses  against  another,  if  the  perpendicular  pres- 
sure is  the  same  the  friction  will  be  the  same.  Therefore, 
large  surfaces  are  used  where  possible,  not  to  reduce  the 
friction,  but  to  reduce  the  wear  and  diminish  the  liability  of 
heating. 

For  instance,  if  the  perpendicular  pressure  between  a 
journal  and  its  bearing  is  10,000  pounds,  and  the  coefficient 
of  friction  is  .2,  the  amount  of  friction  is  10,000  X  .2 
=  2,000  pounds.  Suppose  that  the  area  receiving  the  pres- 
sure is  80  square  inches,  then  the  amount  of  friction  for 
each  square  inch  is  2,000  -f-  80  =  25  pounds. 

If  the  area  receiving  the  pressure  had  been  160  square 
inches,  the  friction  would  have  been  the  same,  that  is,  2,000 
pounds;  but  the  friction  per  square  inch  would  have  been 
2,000  -=-  160  =  12£  pounds,  just  one-half  as  much  as  before, 
and  the  wear  and  liability  to  heat  would  be  one-half  as 
great  also. 

92.  The  values  of  the  coefficient  of  friction  given  in  the 
following  tables  are  average  values  determined  by  General 
Morin,    many    years    ago.     Under    certain    conditions    the 
coefficient  may  be  considerably  less   than  is  given  in  the 
tables;  it  varies  greatly,  but  the  variation  depends  on  so 
many  conditions  and  the  numerous  experiments  that  have 
been  made  have  given  such   contradictory  results  that   no 
definite  rules  have  yet  been   derived  for   determining  the 


§  4  PRINCIPLES  OF  MECHANICS.  35 

exact  values  under  any  condition.  The  student  is,  there- 
fore, advised  to  use  the  values  given  in  the  tables,  except 
where  careful  experiments  have  been  made  that  give  relia- 
ble values  for  the  particular  case  under  consideration.  To 
find  the  force,  in  pounds,  necessary  to  overcome  friction, 
the  coefficient  taken  from  the  table  is  multiplied  by  the 
perpendicular  pressure,  in  pounds,  on  the  surface  consid- 
ered. If  the  force  acts  at  an  angle  to  the  surface,  the 
perpendicular  force  can  be  found  by  resolving  the  given 
force  into  two  components,  one  perpendicular  and  the  other 
parallel  to  the  surface. 

TABLE  I. 


COEFFICIENTS    OF    FRICTION    FOR    PLANE    SURFACES. 

(Reduced  from  M.  Morin's  Data.) 


Description  of  Surfaces 
in  Contact. 

State  of  the  Surfaces. 

Coefficient 
of  Friction. 

Wrought  iron  on  cast  iron. 

Slightly  greasy 

.18 

Wrought  iron  on  bronze.. 

Slightly  greasy 

.18 

Cast  iron  on  cast  iron.  .  .  . 

Slightly  greasy 

.15 

Cast  iron  on  bronze 

Slightly  greasy 

.15 

Bronze  on  bronze  

Drv 

.20 

Bronze  on  cast  iron  

J 

Dry 

.22 

Bronze  on  wrought  iron.  . 

Slightly  greasy 

.16 

Cast  iron,  wrought  iron,  1 

steel,  and  bronze  sli- 

Ordinary lubrication 

ding  on    one   another  \ 

with  lard,  tallow,  and 

.07-.  08 

or   sliding    on    them- 

oil 

selves. 

Cast  iron,  wrought  iron,  ' 

\r 

steel,  and  bronze  sli- 

ding on  one   another   , 

Continuous  lubrication 

.05 

or   sliding    on    them- 

selves. 

PRINCIPLES  OF  MECHANICS. 
TABLE  II. 


COEFFICIENTS   OF   FRICTION   FOB   JOURNAL   FRICTION. 

(Reduced  from  M.  Morin's  Data.) 


Journals. 

Bearings. 

Lubricant. 

Coefficient  of  Friction. 

Ordinary 
Lubrication. 

Continuous 
Lubrication. 

Cast  iron 

Cast  iron 

Lard,  olive  oil, 
tallow 

.07-.  08 

.03-.054 

Cast  iron 

Bronze 

Lard,  olive  oil, 
tallow 

.07-.08 

.03-.054 

Wrought  iron 

Cast  iron 

Lard,  olive  oil, 
tallow 

.07-.  08 

.03.  -054 

Wrought  iron 

Bronze 

Lard,  olive  oil, 
tallow 

.07-.  08 

.03-.  054 

Wrought  iron 

Lignum  vita? 

Lard,  olive  oil 

.11 

Bronze 

Bronze 

Oil 

.10 

Bronze 

Bronze 

Lard 

.09 

93.  In  the  case  of  a  weight  sliding  along  a  horizontal 
plane  surface,  the  pressure  is  equal  to  the  weight.     When 
the    surface    is  inclined,  'the  weight    acts  vertically  down- 
wards, and  the  pressure  perpendicular  to  the  surface  can 
be  found  by  the  principle  of  the  resolution  of  forces.     In 
many  cases  the  pressure  on  the  surfaces  is  due  to  the  com- 
bined action  of  several  forces  that  must  be  combined  into 
one  common  resultant  force. 

94.  The  work  that  must  be   done   in   overcoming  the 
resistance  of  friction  depends  on  the  distance  through  which 
the  resistance  is  overcome.     It  may  be  calculated  by  the 
following  rule: 

Rule  9. — Multiply  the  total  pressure  in  pounds  by  the  dis- 
tance in  feet  and  by  the  coefficient  of  friction. 


§  4  PRINCIPLES  OF  MECHANICS.  37 

Let  W  =•  work  in  foot-pounds; 

f—  coefficient  of  friction; 
/  =  total  pressure  in  pounds ; 
d=  distance  in  feet. 

Then,  W  =  pdf. 

EXAMPLE. — The  average  perpendicular  pressure  on  the  guide  of  a 
steam  engine  due  to  the  force  impelling  the  piston  is  2,500  pounds. 
The  pressure  due  to  the  weight  of  the  crosshead  and  connecting-rod 
is  400  pounds.  The  crosshead  moves  at  the  rate  of  500  feet  per  minute ; 
what  horsepower  is  required  to  overcome  the  friction  on  the  guides  ? 

SOLUTION.— The  total  perpendicular  pressure  is  2,500  +  400  =  2,900 
pounds.  Since  the  lubrication  is  usually  intermittent,  the  coefficient 
of  friction,  for  a  brass  slipper  working  on  a  cast-iron  guide,  may  be 
taken  as  .08.  The  resistance  being  overcome  through  a  distance 
of  500  feet  each  minute,  the  work  done  in  overcoming  friction  is 
2,900  X  500  X  .08  =  116,000  foot-pounds  per  minute.  Then,  the  horse- 
power is  116,000  -=-  33,000  =  3.52  H.  P.,  nearly.  Ans. 

95.  In  the  case  of  a  shaft  rotating  in  a  bearing,  the  dis- 
tance through  which  friction  is  overcome   each  minute  is 
found  by  multiplying  the  circumference  of  the  journal  by 
the  number  of  revolutions  per  minute.     For  a  shaft,  or  any 
other   body  rotating    in    a   bearing,   the  force  required   to 
overcome  friction,  as   calculated  by  multiplying  the   pres- 
sure by  the  coefficient  of  friction,  is  the  force  that  must  be 
applied  at  the  surface  of  the  journal. 

96.  Allowable  Pressures. — It  has  been  found  by  expe- 
rience that  when  the  pressure  per  unit  of  area  exceeds  a 
certain   amount,  the  lubricant  will   be  forced   out  and  the 
bodies  rubbing  on  each  other  will  heat  and,  finally,  seize. 

The  pressures  that  can  safely  be  allowed  on  the  bearings 
of  steam  engines,  on  guides,  thrust  bearings,  crankpins, 
crosshead  pins,  etc.  vary  considerably,  being  dependent  to 
a  large  extent  on  the  character  of  the  workmanship,  the 
degree  of  finish,  the  variation  of  the  pressure,  the  character 
of  the  lubrication,  and  the  quality  of  the  lubricant.  With 
fair  workmanship,  the  following  pressures  per  square  inch 
represent  average  practice  in  steam-engine  work: 
11.  8.  I.— IS 


PRINCIPLES  OF  MECHANICS. 
TABLE    III. 


PRESSURES  PER  SQUARE   INCH  ALLOWABLE  IN 
STEAM-ENGINE  \VORK. 


Engine  Bearing. 

Slow-Speed 
Stationary  Engines. 

Pounds. 

High-Speed 
Stationary  and 
Marine    Engines. 

Pounds. 

Crankpins    iron    .... 

800) 

Crankpins   steel  

1  200  f 

400  to  600 

WVistpin 

1  200 

000  to  800 

Main   bearings  
Guides 

200 
100 

200 
100 

Thrust  bearings   ...    . 

60 

97.  For  crankpins,  wristpins,  and  guides,  the  allowable 
pressures  given  represent  the  pressures  corresponding  to  the 
maximum  load,  which  in  the  case  of  a  wristpin  and  crankpin 
occurs  when  the  crank,  connecting-rod,  and  piston  rod  are 
in  a  straight  line,  and  in  the  case  of  guides,  when  the  con- 
necting-rod and  crank  are  at  right  angles  to  each  other.  In 
the  case  of  pins  and  journals,  the  area  to  be  considered  in 
calculating  the  pressure  on  the  bearing  is  the  projected 
area,  which  is  found  by  multiplying  the  length  of  the 
journal  by  its  diameter. 


EXAMPLES    FOR   PRACTICE. 

1.  A  body  weighs  90  pounds ;  what  is  its  mass  ?      Ans.  2.799,  nearly. 

2.  What  force  will  be  required  to  accelerate  a  body  at  the  rate  of 
2  feet  per  second,  the  body  weighing  450  pounds,  and  the  frictional 
resistances  being  equal  to  10  per  cent,  of  the  weight  of  the  body  ? 

Ans.  72.99  lb.,  nearly. 
8.     What  work  is  done  in  raising  950  pounds  17  feet  ? 

Ans.  16,150  ft.-lb. 

4.     If  an  engine  does  205,000  foot-pounds  of  work  per  minute,  what 
is  its  horsepower  ?  Ans.  6.21  H.  P.,  nearly. 


PRINCIPLES  OF  MECHANICS. 


39 


5.  What  is  the  kinetic  energy  of  a  shell  fired  from  a  cannon  with  a 
velocity  of  1,800  feet  per  second,  the  shell  weighing  1,000  pounds? 

Ans.  50,373,135  ft.-lb.,  nearly. 

6.  Taking  the  coefficient  of  friction  at  .15,  what  horsepower  will  be 
required  to  pull  100  pounds  at  a  uniform  speed  of  5  feet  per  second 
along  a  level  surface  ?  Ans.  .136  H.  P. 


CENTER    OF    GRAVITY. 

98.  The  center  of  gravity  of  a  body  is  tJiat  point  at  which 
the  body  may  be  balanced,  or  it  is  the  point  at  which  the  whole 
weight  of  a  body  may  be  considered  as  concentrated. 

This  point  is  not  always  in  the  body;  in  the  case  of  a 
horseshoe  or  a  ring  it  lies  outside  of  the  substance  of,  but 
within  the  space  enclosed  by,  the  body. 

In  a  moving  body,  the  line  described  by  its  center  of 
gravity  is  always  taken  as  the  path  of  the  body.  In  finding 
the  distance  that  a  body  has  moved,  the  distance  that  the 
center  of  gravity  has 
moved  is  taken. 

The  definition  of  the 
center  of  gravity  of  a 
body  may  be  applied 
to  a  system  of  bodies 
if  they  are  considered 
as  being  connected 

at     their    centers    of 

FIG.  9. 
gravity. 

If  w  and  W,  Fig.  9,  are  two  bodies  of  known  weight,  their 
center  of  gravity  will  be  at  C.  This  point  C  may  be  readily 
determined  as  follows: 

Rule  1O. —  The  distance  of  tlie  common  center  of  gravity- 
front  the  center  of  gravity  of  the  large  weight  is  equal  to  the 
weight  of  the  smaller  body  multiplied  by  the  distance  between 
the  centers  of  gravity  of  t/ie  two  bodies,  and  this  product 
divided  by  the  sum  of  the  weights  of  the  two  bodies. 

EXAMPLE. — In  Fig.  9,  w  =  10  pounds,  IV  —  30  pounds,  and  the  dis- 
tance between  their  centers  of  gravity  is  36  inches;  where  is  the  center 
of  gravity  of  both  bodies  situated  ? 


40 


PRINCIPLES  OF  MECHANICS. 


SOLUTION.— Applying   the   rule,    10  X  36  =  360.     10  +  30  =  40.     360 
-s-  40  =  9  in.,  distance  of  center  of  gravity  from  center  of  large  weight. 

Ans. 

99.  It  is  now  very  easy  to  extend  this  principle  to  find 
the  center  of  gravity  of  any  number  of  bodies,  when  their 
weights  and  the  distances  apart  of  their  centers  of  gravity 
are  known,  by  the  following  rule : 

Rule  11. — Find  the 
center  of  gravity  of  two 
of  the  bodies,  as  Wl 
and  W<  in  Fig.  10.  As- 
sume that  the  weight 
of  both  bodies  is  concen- 
trated at  Cj,  and  find  the 
center  of  gravity  of  this 
combined  weight  at  Ct  and 
the  weight  of  W^.  Let  it 
be  at  C^;  then  find  the 
center  of  gravity  of  the 
combined  weights  of  W^, 
W4,  W^  (concentrated  at 

C^),  and  W3.     Let  it  be  at  C ;  then  C  will  be  the  center  of 

gravity  of  the  four  bodies. 

100.  To   find   the   center  of   gravity  of  any  parallel- 
ogram: 

Rule  12. — Draw  the  two  diagonals,   Fig.    11,    and  their 
point  of  intersection  C  will  be  the  center  of  gravity. 


1O1.     To    find   the  center  of  gravity  of  a   triangle,  as 
ABC,  Fig.  12: 


§  4  PRINCIPLES  OF  MECHANICS.  41 

Rule  13. — From  any  vertex,  as  A,  draw  a  line  to  the  mid- 
dle point  D  of  the  opposite  side  B  C.  From  one  of  the  other 
vertexes,  as  C,  draw  a  line  to  F,  the  middle  point  of  the 
opposite  side  A  B ;  the  point  of  intersection  O  of  these  two 
lines  is  the  center  of  gravity. 

It  is  also  true  that  the  distance  D  O  —  \  D  A  and  that 
F  O  •=•  \  F  C\  the  center  of  gravity  could  have  been  found 
by  drawing  from  any  vertex  a  line  to  the  middle  point  of 
the  opposite  side  and  measuring  back  from  that  side  £  of  the 
length  of  the  line. 

The  center  of  gravity  of  any  regular  plane  figure  is  the 
same  as  the  center  of  the  inscribed  or  circumscribed  circle. 

1O2.  To  find  the  center  of  gravity  of  any  Irregular 
plane  figure,  but  of  uniform  thickness  throughout,  divide 


one  of  the  parallel  surfaces  into  triangles,  parallelograms, 
circles,  ellipses,  etc.,  according  to  the  shape  of  the  figure; 
find  the  area  and  center  of  gravity  of  each  part  separately, 
and  combine  the  centers  of  gravity  thus  found  in  the  same 


42  PRINCIPLES  OF  MECHANICS.  §  4 

manner  as  in  rule   11,  in  this  case,  however,  dealing  with 
the  area  of  each  part  instead  of  its  weight.      See  Fig.  13. 

1O3.  Center  of  Gravity  of  a  Solid. — In  a  body  free 
to  move,  the  center  of  gravity  will  lie  in  a  vertical  plumb- 
line  drawn  through  the  point  of  support.  Therefore,  to 


FIG.  14. 

find  the  position  of  the  center  of  gravity  of  an  irregular 
solid,  as  the  crank,  Fig.  14,  suspend  it  at  some  point,  as  B, 
so  that  it  will  move  freely.  Drop  a  plumb-line  from  the 
point  of  suspension,  and  mark  its  direction.  Suspend 
the  body  at  another  point,  as  A,  and  repeat  the  process. 
The  intersection  C  of  the  two  lines  will  be  directly  over 
the  center  of  gravity. 

It  is  often  desired  to  find  the  horizontal  distance  of  the 
center  of  gravity  from  a  given  point  of  the  body.  In  many 
cases  this  can  readily  be  done  by  balancing  the  body  on  a 
knife  edge,  and  then  measuring,  horizontally,  the  distance 
between  the  knife  edge  and  the  given  point. 

Since  the  position  of  the  center  of  gravity  depends  wholly 
on  the  shape  and  weight  of  a  body,  it  may  be  without  the 
body,  as  in  the  case  of  a  circular  ring,  whose  center  of  gravity 
is  the  same  as  the  center  of  the  circumference  of  the  ring. 

By  considering  the  symmetry  of  form,  the  position  of  the 


§  4  PRINCIPLES  OF  MECHANICS.  43 

center  of  gravity  of  homogeneous  solids  may  often  be  deter- 
mined without  analysis,  or  it  may  be  limited  to  a  certain 
plane,  line,  or  point.  Thus,  the  center  of  gravity  of  a  sphere, 
or  any  other  regular  body,  is  situated  at  its  center;  of  a 
cylinder,  in  the  middle  of  its  axis;  of  a  thin  plate  having 
the  form  of  a  circle  or  regular  polygon,  in  the  center  of  the 
figure;  of  a  straight  wire  of  uniform  cross-section,  in  the 
middle  of  its  length. 

EXAMPLES    FOR   PRACTICE. 

1.  A  spherical  shell  has  a  wrought-iron  handle  attached  to  it.     The 
shell  is  10  inches  in  diameter  and  weighs  20  pounds.     The  handle  is 
1|  inches  in  diameter,  and  the  distance  from  the  center  of  the  shell  to 
the  end  of  the  handle  is  4  feet.    Where  is  the  center  of  gravity  ?    Take 
the  weight  of  a  cubic  inch  of  wrought  iron  as  .278  pound. 

Ans.  13.612  in.  from  center  of  shell. 

2.  The  distance  between  the  centers  of  two  bodies  is  51  inches. 
The  weights  of  the  bodies  being  20  and  73  pounds,  where  is  the  center 
of  gravity  ?  Ans.  10.968  in.  from  the  center  of  the  larger  weight. 

3.  Weights  of  5,  9,  and  12  pounds  lie  in  one  straight  line  in  the 
order   named.     Distance   from   the  5-pound  weight   to   the    9-pound 
weight  is  22  inches,  and  from  the  9-pound  weight  to  the  12-pound 
weight  is  18  inches.     Where  is  the  center  of  gravity  ? 

Ans.  13.923  in.  from  12-lb.  weight. 


CF/NTRIFTJGAI,    FORCE. 

1O4.  If  a  body  is  fastened  to  a  string  and  whirled,  so 
as  to  give  it  a  circular  motion,  there  will  be  a  pull  on  the 
string  that  will  be  greater  or  less,  according  as  the  velocity 
increases  or  decreases.  The  cause  of  this  pull  on  the  string 
will  now  be  explained. 

Suppose  that  the  body  is  revolved  horizontally,  so  that  the 
action  of  gravity  upon  it  will  always  be 
the  same.  According  to  the  first  law  of 
motion,  a  body  put  in  motion  tends  to 
move  in  a  straight  line  unless  acted  upon 
by  some  other  force,  causing  a  change  in 
the  direction.  When  the  body  moves  in 
a  circle,  the  force  that  causes  it  to  move  FIG.  15. 


44  PRINCIPLES  OF  MECHANICS.  §4 

in  a  circle  instead  of  in  a  straight  line  is  exactly  equal  to  the 
tension  of  the  string.  If  the  string  were  cut,  the  pulling 
force  that  drew  it  away  from  the  straight  line  would  be 
removed,  and  the  body  would  then  "fly  off  at  a  tangent"; 
that  is,  it  would  move  in  a  straight  line  tangent  to  the 
circle,  as  shown  in  Fig.  15. 

Since,  according  to  the  third  law  of  motion,  every  action 
has  an  equal  and  opposite  reaction,  we  call  the  force  that 
acts  as  an  equal  and  opposite  force  to  the  pull  of  the  string 
the  centrifugal  force,  and  it  acts  away  from  the  center 
of  motion. 

105.  The  other  force,  or  tension,  of  the  string  is  called 
the   centripetal   force,  and  it    acts  toward  the  center  of 
motion.     It  is  evident  that  these  two  forces,  acting  in  oppo- 
site directions,   tend    to  pull  the  string  apart,   and,   if   the 
velocity  be  increased  sufficiently,  the  string  will  break.     It 
is  also  evident  that  no  body  can  revolve  without  generating 
centrifugal  force. 

106.  To   Find   the    Centrifugal   Force   of  Any   Re- 
volving   Body. — The     value     of     the    centrifugal     force, 
expressed  in  pounds,  of  any  revolving  body  is  calculated  by 
the  following  rule: 

Rule  14. —  The  centrifugal  force  equals  the  continued 
product  of  .00034,  the  weight  of  the  body  in  pounds,  the  radius 
in  feet  (taken  as  the  distance  between  the  center  of  gravity  of 
the  body  and  the  center  about  which  it  revolves),  and  the 
square  of  the  number  of  revolutions  per  minute. 

Let  F '  =  centrifugal  force  in  pounds; 

W=  weight  of  revolving  body  in  pounds; 

R  =  radius  in  feet  of   circle    described  by  center  of 

gravity  of  revolving  body; 
Ar=  revolutions  per  minute  of  revolving  body. 

Then,  F=  .00034  WRN*. 

EXAMPLE. — What  is  the  tension  in  the  string  of  Fig.  15,  if  the  ball 
weighs  2  pounds  and  is  revolved  around  at  the  rate  of  500  revolutions 
per  minute  ?  The  string  is  2  feet  long. 


§  4  PRINCIPLES   OF   MECHANICS.  45 

SOLUTION.  —  Applying  the  rule  just  given,  we  get 

F=  .00034  X  2  x  2  X  500-  =  340  Ib.     Ans. 

107.  In  flywheels,  belt  wheels,  and  pulleys  the  centrif- 
ugal force  tends  to  tear  the  rim  asunder;  this  tendency  is 
resisted  by  the  tenacity  of  the  material  of  which  the  wheel 
is  composed.      Since  the  centrifugal  force  increases  as  the 
square  of  the  number  of  revolutions,  it  will  be  seen  that  an 
apparently  slight  increase  in  the  number  of  revolutions  per 
minute  may  be  sufficient  to  burst  the  wheel. 

108.  For  solid  cast-iron  wheels  and  for  built-up  wheels 
of  cast  iron  where  the  strength  of  the  joint  is  equal  to  the 
strength  of  the  rim,  the  greatest  number  of  revolutions  per 
minute  that  practice  has  indicated  to  be  safe  may  be  found 
by  the  following  rule  : 

Rule  15.  —  Divide  1,930  by  the  diameter  of  the  wheel. 


Or. 


where          d  =  diameter  of  the  wheel  in  feet; 

N=  number  of  revolutions  per  minute. 

EXAMPLE.  —  What  is  the  maximum  number  of  revolutions  allowable 
for  a  cast-iron  flywheel  27  feet  6  inches  in  diameter  ? 
SOLUTION.  —  Applying  rule  15,  we  get 

N  =    '        =  70  rev.  per  min.     Ans. 


EQUILIBRIUM. 

1O9.  When  a  body  is  at  rest,  all  the  forces  that  act  upon 
it  balance  one  another  and  are  said  to  be  in  equilibrium. 
The  most  important  force  to  be  considered  is  the  attraction 
of  the  earth,  which  acts  upon  every  particle  of  a  body. 
There  are  three  states  of  equilibrium :  stable,  unstable,  and 
neutral. 

HO.  A  body  is  in  stable  equilibrium  when,  if  slightly 
rotated  about  its  support  in  such  manner  as  to  change  the 


46  PRINCIPLES  OF  MECHANICS.  §  4 

elevation  of  its  center  of  gravity,  it  tends  to  return  to  its 
position  of  rest.  Examples  of  bodies  in  a  state  of  stable 
equilibrium  are  a  cube  resting  on  one  of  its  sides,  a  cone 
resting  on  its  base,  a  pendulum,  etc.  A  body  can  only  be 
in  stable  equilibrium  when  a  rotation  about  its  support  raises 
the  center  of  gravity. 

111.  A  body  is  in  unstable  equilibrium  when  a  rota- 
tion about  its  support,  so  as  to  change  the  elevation  of  its 
center  of  gravity,  tends  to  make  it  fall  farther  from  its  posi- 
tion of  rest.  Examples  of  bodies  in  unstable  equilibrium 
are  a  cube  balanced  on  one  of  its  edges,  a  cone  standing 
upon  its  point,  an  egg  balanced  upon  its  end,  etc.  When 
a  body  is  in  unstable  equilibrium,  any  rotation,  no  matter 
how  slight,  tends  to  lower  its  center  of  gravity. 


A  body  is  in  neutral  equilibrium  when  a  rota- 
tion about  its  support  does  not  change  the  elevation  of  its 
center  of  gravity.  Examples  of  bodies  in  neutral  equilib- 
rium are  a  sphere  of  uniform  density  and  a  cone  resting  on 
its  side. 

113.  A  vertical  line  drawn  through  the  center  of  grav- 
ity of  a  body  is  called  the  line  of  'direction.  So  long  as  the 
line  of  direction  falls  within  the  base,  the  body  will  stand; 
when  the  line  of  direction  falls  outside  of  the  base,  the  body 
will  fall. 

B 


Let  A  CB,  Fig.  16,  be  a  cylinder  whose  base  is  oblique  to 
the  center  line  B  O  D,  and  let  O  be  the  center  of  gravity  of 


§  4  PRINCIPLES  OF  MECHANICS.  4? 

the  cylinder.  So  long  as  the  vertical  line  drawn  through  O 
falls  between  A  and  C,  the  cylinder  will  stand,  but  the 
instant  it  falls  without  the  base,  the  cylinder  will  fall. 

114.     The  center  of  gravity  of  a  body  has  a  tendency  to 
always  seek  the  lowest  possible  position. 


MACHINE    ELEMENTS. 


THE    LEVER,  WHEEL,  AND  AXLE. 


FUNDAMENTAL    PRINCIPLES. 

1.     A  lever  is  a   bar  capable  of  being  turned  about  a 
pivot,  or  point,  as  in  Figs.  1,  2,  and  3. 


r 


The  object  IV to  be  lifted  is  called  the  weight;  the  force 
is  represented  by  P* ;  and  the  point,  or  pivot  F,  is  called  the 
fulcrum. 


*  The  force  applied  to  a  lever,  screw,  wheel  and  axle,  or  any  similar 
machine  element  in  order  to  raise  a  given  weight,  was  formerly  called 
the  power,  and  the  arm  of  a  lever  to  which  the  force  is  applied  was 
called  the  power  arm  ;  at  present,  however,  the  term  power  is  uni- 
versally used  to  represent  the  rate  at  which  work  is  done,  and,  hence, 
its  application  to  those  cases  where  a  simple  force  is  meant  often  leads 
to  a  serious  confusion  of  ideas  regarding  the  relation  between  force, 
work,  and  power.  To  prevent  this  confusion,  we  will  use  the  word 
power  only  in  accordance  with  the  definition  given. 

§5 

For  notice  of  the  copyright,  see  page  immediately  following  the  title  page. 


2  MACHINE  ELEMENTS.  §  5 

2.  That  part  of  the  lever  F  b  between  the  fulcrum  and 
the  weight   is  called  the  weight  arm,   and   the    part  F  c 
between  the  fulcrum  and  the  force  is  called  the  force  arm. 

In  order  that  the  lever  will  be  in  equilibrium  (balance), 
the  force  multiplied  by  the  force  arm  must  equal  tlie  weight 
multiplied  by  the  weight  arm ;  that  is,  P  X  F  c  must 
equal  W  X  F  b. 

3.  If  F  is  taken  as  the  center  of  a  circle,  and  arcs  are 
described  through  b  and  £,  it  will  be  seen  that  if  the  weight 
arm  is  moved  through  a  certain  angle,  the  force  arm  will 
move  through  the  same  angle.     Since  in  the  same  or  equal 
angles  the  lengths  of  the  arcs  are  proportional  to  the  radii 
with  which  they  were  described,  it  is  seen  that  the  force 
arm  is  proportional  to  the  distance  through  which  the  force 
acts,   and  the  weight  arm  is  proportional  to  the  distance 
through  which  the  weight  moves.     Hence,  instead  of  wri- 
ting Px  F  c—  IV  X  F  b,  we  might  have  written  P  X  (dis- 
tance   through   which   P  acts)  =   W  X   (distance    through 
which  Amoves).     This  is  the  general  law  of  all  machines, 
and  can  be  applied  to  any  mechanism  from  the  simple  lever 
up  to  the  most  complicated  arrangement.     When  stated  in 
the  form  of  a  rule  it  is  as  follows: 

Rule  1. —  The  force  multiplied  by  the  distance  through 
which  it  acts  equals  the  weight  multiplied  by  the  distance 
through  which  it  moves. 

4.  In  the  above  rule,  it  will  be  noticed  that  there  are 
four  requirements  necessary  for  a  complete  knowledge  of 
the  lever,  viz. :  the  force,   the  weight,  the   force   arm,   or 
distance  through  which  the  force  acts,  and  the  weight  arm, 
or  distance  through  which  the  weight  moves.     If  any  three 
are  given,  the  fourth  may  be  found  by  letting  x  represent 
the  requirement  that  is  to  be  found,  and  multiplying  the 
force  by  the  force  arm  and  the  weight  by  the  weight  arm ; 
then,  dividing  the  product  of  the  two  known  numbers  by 
the  number  by  which  x  is  multiplied,  the  result  will  be  the 
requirement  that  is  to  be  found. 


§  5  MACHINE  ELEMENTS.  3 

EXAMPLE.— If  the  weight  arm  of  a  lever  is  6  inches  long  and  the 
force  arm  is  4  feet  long,  how  great  a  weight  can  be  raised  by  a  force  of 
20  pounds  at  the  end  of  the  force  arm  ? 

SOLUTION. — In  this  example,  the  weight  is  unknown;  hence,  repre- 
senting it  by  x,  we  have,  after  reducing  the  4  feet  to  inches,  20  X  48 
=  960  =  force  multiplied  by  the  force  arm,  and  x  X  6  =  weight  mul- 
tiplied by  the  weight  arm.  Dividing  the  960  by  6,  the  result  is 
160  pounds,  the  weight.  Ans. 

5.  If  the  distance  through  which  the  force  acted  or  the 
weight  had  moved  had  been  given  instead  of  the  force  arm  or 
weight  arm,  and  it  were  required  to  find  the  force  or  weight, 
the  process  would  have  been    exactly  the  same,  using  the 
given  distance  instead  of  the  force  arm  or  weight  arm. 

EXAMPLE.— If,  in  the  above  example,  the  weight  had  moved 
2^  inches,  through  what  distance  would  the  force  have  acted  ? 

SOLUTION. — In  this  example,  the  distance  through  which  the  force 
acts  is  required.  Let  x  represent  the  distance.  Then,  20  X  x  =  dis- 
tance multiplied  by  force,  and  2£  X  160  =  400  =  distance  multiplied  by 
the  weight.  Hence,  x  =  4^°-  =  20  inches  =  distance  through  which  the 
force  arm  moves.  Ans. 

The  ratio  between  the  weight  and  the  force  is  160  H-  20  =  8.  The 
ratio  between  the  distance  through  which  the  weight  moves  and  the 
distance  through  which  the  force  acts  is  2£  -=-  20  =  \.  This  shows  that 
while  a  force  of  1  pound  can  raise  a  weight  of  8  pounds,  the  1-pound 
weight  must  move  through  8  times  the  distance  that  the  8-pound 
weight  does.  It  will  also  be  noticed  that  the  ratio  of  the  lengths  of 
the  two  arms  of  the  lever  is  also  8,  since  48  -*-  6  =  8. 

6.  The  law  that  governs  the  straight  lever  also  governs 
the  bent  lever,  but  care  must  be  taken  to  determine  the  true 
lengths  of  the  lever  arms,  which  are,  in  every  case,  the  per- 
pendicular distances  from  the  fulcrum  to  the  line  of  direction 
of  the  weight  or  force. 

Thus,  in  Figs.  4,  5,  6,  and  7,  F  c  in  each  case  represents 
the  force  arm  and  F  b  the  weight  arm. 

7.  A    compound    lever    is    a   series   of   single   levers 
arranged  in  such  a  manner  that  when  a  force  is  applied  to 
the  first  it  is  communicated  to  the  second,  and  from  that  to 
the  third,  and  so  on. 


4  MACHINE  ELEMENTS.  §  5 

Fig.  8  shows  a  compound  lever.  It  will  be  seen  that  when 
a  force  is  applied  to  the  first  lever  at  P  it  will  be  communi- 
cated to  the  second  lever  at  /*,  from  this  to  the  third  lever 
at  P,  and  thus  raise  the  weight  W. 


FIG.  5. 


The  weight  that  the  force  applied  to  the  first  lever  could 
raise  acts  as  the  force  of  the  second,  and  the  weight  that  this 
force  could  raise  through  the  second  lever  acts  as  the  force 


FIG.  0.  FIG.  7. 

of  the  third  lever,  and  so  on,  no  matter  how  many  single 
levers  make  up  the  compound  lever. 

In  this  case,  as  in  every  other,  the  force  multiplied  by  the 


distance  through  which  it  acts  equals  the  weight  multiplied 
by  the  distance  through  which  it  moves. 


§  5  MACHINE  ELEMENTS.  5 

Hence,  if  we  move  the  P  end  of  the  lever,  say  4  inches, 
and  the  end  carrying  the  weight  W  moves  |  inch,  we  know 
that  the  ratio  between  P  and  W  is  the  same  as  the  ratio 
between  \  and  4;  that  is,  1  to  20,  and,  hence,  that  10  pounds 
at  /'will  balance  200  pounds  at  W,  without  measuring  the 
lengths  of  the  different  lever  arms.  If  the  lengths  of  the 
lever  arms  are  known,  the  ratio  between  P  and  J^may  be 
readily  obtained  from  the  following  rule: 

Rule  2. —  The  continued  product  of  the  force  and  each  force 
arm  equals  the  continued  product  of  the  weight  and  each  weight 
arm. 

EXAMPLE.— If,  in  Fig.  8,  P  F=  24  inches,  18  inches,  and  30  inches, 
respectively,  and  W  F=.§  inches,  6  inches,  and  18  inches,  respectively, 
how  great  a  force  at  P  will  it  require  to  raise  1,000  pounds  at  W? 
What  is  the  ratio  between  W  and  P  ? 

SOLUTION.  —  Let  x  represent  the  force ;  then,  x  X  24  X  18  X  30 
=  12,960  X  x  =  continued  product  of  the  force  and  each  force  arm. 
1,000  X  6  X  6  X  18  =  648,000  =  continued  product  of  the  weight  and 
each  weight  arm;  and,  since  12,960  X  x  =  648,000, 

648,000 

—  =  50  Ib.  =  the  force.     Ans. 


1,000  -H  50  =  20  =  ratio  between  fFand  P.     Ans. 

8.     The   wheel   and  axle   consists  of  two  cylinders  of 
different    diameters   rigidly   connected,    so   that    they    turn 


FIG.  9. 


//.  s.    L—U 


6  MACHINE  ELEMENTS.  §  5 

together  about  a  common  axis,  as  in  Fig.  9.  Then,  as 
before,  P  X  distance  through  which  it  acts  =  W  X  distance 
through  which  it  moves ;  and,  since  these  distances  are  pro- 
portional to  the  radii  of  the  force  cylinder  and  weight  cylin- 
der, P  X  Fc  =  W  X  Fb. 

It  is  not  necessary  that  an  entire  wheel  be  used ;  an  arm, 
projection,    radius,    or    anything 

P    \\  that  the  force  causes  to  revolve 

in  a  circle,  may  be  considered  as 
the  wheel.  Consequently,  if  it  is 
desired  to  hoist  a  weight  with  a 
windlass,  Fig.  10,  the  force  is  ap- 
plied to  the  handle  of  the  crank, 
and  the  distance  between  the 
center  line  of  the  crank-handle 

and  the  axis  of  the  drum  corresponds  to  the  radius  of  the 
wheel. 

EXAMPLE.— If  the  distance  between  the  center  line  of  the  handle 
and  the  axis  of  the  drum  in  Fig.  10  is  18  inches  and  the  diameter  of 
the  drum  is  6  inches,  what  force  will  be  required  at  P  to  raise  a  load 
of  300  pounds  ? 

SOLUTION. —    P  x  (18  X  2)  =  300  x  6,  or  P  =  50.     Ans. 


EXAMPLES   FOR  PRACTICE. 

1.  The  lever  of  a  safety  valve  is  of  the  form  shown  in  Fig.  1,  where 
the  force  is  applied  at  a  point  between  the  fulcrum  and  the  weight 
lifted.     If  the  distance  from  the  fulcrum  to  the  valve  is  5£  inches  and 
from  the  fulcrum  to  the  weight  is  42  inches,  what  total  force  is  neces- 
sary to  raise  the  valve,  the  weight  being  78  pounds  and  the  weight  of 
valve  and  lever  being  neglected  ?  Ans.  595.64  Ib. 

2.  If,  in  Fig.  8,  P  F—  10  inches,  12  inches,  14  inches,  and  16  inches, 
respectively,  and  W  F=  2  inches,  3  inches,  4  inches,  and  5   inches, 
respectively,  (a)  how  great  a  weight  can  a  force  of  20  pounds  raise  ? 
(b)  What  is  the  ratio  between  W  and  P  ?    (c)  If  P  moves  4  inches, 
how  far  will  W  move  ?  i  (a)    4,480  Ib. 

Ans.  \  (6)     224. 
•(')     A  in- 


§  5  MACHINE  ELEMENTS.  7 

3.  A  windlass  is  used  to  hoist  a  weight.  If  the  diameter  of  the 
drum  on  which  the  rope  winds  is  4  inches,  and  the  distance  from  the 
center  of  the  handle  to  the  axis  of  the  drum  is  14  inches,  how  great  a 
weight  can  a  force  of  32  pounds  applied  to  the  handle  raise  ? 

Ans.  234  Ib. 


PTJI.LEYS. 

9.  Pulleys  for  the  transmission  of  power  by  belts  may  be 
divided  into  two  principal  classes:  (1)  The  solid  pulley 
shown  in  Fig.  11,  in  which  the  hub,  arms,  and  rim  are  one 
entire  casting.  (2)  The  split  pulley  shown  in  Fig.  12,  which 
is  cast  in  halves. 


The  latter  style  of  pulley  is  more  readily  placed  on  and 
removed  from  the  shaft  than  the  solid  pulley.  Pulleys  are 
generally  cast  in  halves  or  parts  when  they  are  more  than 
6  feet  in  diameter;  this  is  done  on  account  of  the  shrink- 
age strain  in  large  pulley  castings,  which  renders  them 
liable  to  crack  as  a  result  of  the  unequal  cooling  of  the 
metal. 


8 


MACHINE  ELEMENTS. 


10.  Of  late  years,  wooden  pulleys  have  come  into  exten- 
sive use.     They  are  built  of  segments  securely  glued  together, 
maple  being  the  wood  used.     Wooden  split  pulleys  can  be 
procured  that  are  fitted  with  removable  bushings,  thus  allow- 
ing the  same  pulley  to  be  readily  adapted  to  various  diam- 
eters of  shafting.     They  are  somewhat  lighter  than  cast-iron 
pulleys. 

11.  Crowning  Pulley  Faces. — In  Fig.  13,  suppose  the 
shafts  a  and  b  to  be  parallel.     Let  the  pulley  on  the  shaft  a 

be  cone-shaped,  as  shown.  The  right-hand 
side  of  the  belt  will  be  pulled  ahead  more 
rapidly  than  the  left-hand  side,  because  of  the 
greater  diameter  and,  consequently,  greater 
speed  of  the  right-hand  end  of  the  pulley.  It 
has  been  observed  that  in  this  case  the  belt 
will  leave  its  normal  position,  which  is  indi- 
cated by  the  dotted  lines,  and  climb  toward 
the  part  of  the  pulley  that  has  the  largest 
diameter,  as  shown  in  the  illustration.  This 
tendency  of  the  belt  to  climb  toward  the  high 
side  is  taken  advantage  of  to  make  the  belt 
stay  on  a  pulley.  Suppose  the  pulley  on  the 
shaft  a  is  replaced  with  one  formed  of  two 
equal  cones,  with  the  large  diameter  in  the 
KIG>  13>  center.  Then,  each  side  of  the  belt  will  tend 

to  climb  toward  the  highest  point,  and  the  consequence  is 

that  the  belt  will  stay  in  the  center  of  the  pulley. 

A  pulley  with  its  surface  formed  in  this  way  is  said  to  be 

crowned.     The  surface  need  not  necessarily  represent  the 

frustums  of  two  cones;  it  may  simply 

be  curved,  as  shown  in  Fig.  14.     It  is 

only  required  that  the  pulley  be  larger 

in  diameter  in  the  center. 

As  to  the  proper  amount  of  crowning 

necessary,  the  practice  of  the  makers  FIO.M. 

of  pulleys  differs  considerably ;  usually,  though,  it  is  from 

•£$  to  £  inch  per  foot  of  width  of  the  pulley  face. 


MACHINE  ELEMENTS. 


9 


12.  Balancing  Pulleys. — All  pulleys  that  rotate  at  high 
speeds  should  be  balanced.  If  they  are  not,  the  centrifugal 
force  that  is  gener- 
ated by  the  rota- 
tion of  the  pulley, 
is  greater  on  one 
side  than  on  the 
other  and  will  cause 
the  pulley  shaft  to 
vibrate  and  shake. 
Pulleys  are  usually 
balanced  in  the 
manner  shown  in  FIG.  is. 

Fig.  15.  A  closely  fitting  arbor,  which  is  simply  a  truly 
cylindrical  piece  of  iron  or  steel,  is  driven  into  the  bore  of 
the  pulley,  which  is  then  placed  on  the  so-called  "  balancing 
ways."  These  are  two  planed  iron  or  steel  bars,  preferably 
planed  to  a  knife  edge.  These  bars  are  placed  on  conve- 
nient supports  far  enough  apart  to  allow  the  pulley  to  go 
between  them.  The  bars  should  be  carefully  leveled  with  a 
spirit  level,  so  that  both  bars  are  in  the  same  horizontal 
plane.  When  the  arbor  rests  on  the  ways  and  the  pulley  is 
slightly  rotated,  the  latter  will  quickly  come  to  rest  with 
the  heavy  part  downwards.  Now,  either  some  metal  must 
be  removed  from  the  heavy  part  or  some  weight  added 
to  the  light  part.  For  small  pulleys,  a  convenient  substance 
to  use  for  finding  the  proper  location  and  weight  of  the 
counterweight  is  ordinary  glazier's  putty.  By  repeated 
trials  the  proper  weight  of  the  counterweight  is  found,  and 
then  a  mass  of  metal  of  convenient  shape  equal  in  weight  to 
the  putty  is  fastened  to  the  light  part  of  the  pulley  in  what- 
ever way  is  safe  and  convenient.  The  proper  weight  and 
location  of  the  counterweight  will  have  been  obtained  when 
the  pulley  will  be  at  rest  on  the  balancing  ways  in  any  posi- 
tion in  which  it  is  put.  In  other  words,  the  pulley  is 
balanced  when  it  is  in  neutral  equilibrium.  This  balance  is 
called  a  standing  balance.  The  method  just  explained 
answers  very  well  for  pulleys  that  are  narrow  in  comparison 


10  MACHINE  ELEMENTS.  §  5 

to  their  diameter.  When  pulleys  with  a  wide  face  are  run 
at  a  high  speed,  it  is  often  found  that  serious  vibrations  are 
set  up  even  when  they  are  in  perfect  standing  balance,  thus 
showing  that  they  do  not  possess  the  so-called  running 
balance.  No  method  has  yet  been  found  that  will  insure  a 
running  balance  at  all  speeds,  nor  has  any  method  become 
known  by  which  it  can  be  discovered  directly  where  to  apply 
the  counterweight.  The  usual  remedy  for  pulleys  not  pos- 
sessing a  running  balance  is  to  turn  carefully  the  inside  of 
the  rim  to  run  true  with  the  outside  of  the  pulley.  Absence 
of  running  balance  is,  in  all  cases,  due  to  an  unequal  distri- 
bution of  metal  in  reference  to  the  axis  of  the  shaft,  this 
unequal  distribution  being  due  to  lack  of  homogeneity  of 
the  metal,  to  poor  foundry  work,  or  to  poor  lathe  work. 

13.  Pulleys  should  run  true  in  order  that  the  strain,  or 
tension,  of  the  belt  will  be  equal  at  all  parts  of  the  revolu- 
tion,   thus   making   the    transmitting    power    equal.     The 
smoother  the  surface  of  a  pulley,  the  greater  is  its  driving 
power. 

The  transmitting  power  of  a  pulley  can  be  increased  by 
covering  its  face  with  a  leather  or  rubber  band;  this  in- 
creases the  driving  power  about  one-quarter. 

14.  Relation  Between  Speeds  of  Drivers  and  Driven 
Pulleys. — The   pulley  that  imparts   motion  to  the  belt   is 
called  the  driver ;  that  which  receives  the  motion  is  called 
the  driven. 

The  revolutions  of  any  two  pulleys  over  which  a  belt  is 
run  vary  in  an  inverse  proportion  to  their  diameters;  conse- 
quently, if  a  pulley  20  inches  in  diameter  is  driven  by  one 
10  inches  in  diameter,  the  20-inch  pulley  will  make  1  revolu- 
tion while  the  10-inch  pulley  makes  2  revolutions,  or  they 
are  in  the  ratio  of  2  to  1.  From  this  fact,  the  following 
formulas  have  been  deduced : 

Let       D  =  diameter  of  the  driver; 
d  =  diameter  of  the  driven ; 
N=  number  of  revolutions  of  the  driver; 
n  =  number  of  revolutions  of  the  driven. 


§  5  MACHINE  ELEMENTS.  11 

NOTE.  —  The  words  revolutions  per  minute  are  frequently  abbre- 
viated to  R.  P.  M. 

15.  To  find  the  diameter  of  the  driving  pulley  when  the 
diameter  of  the  driven  pulley  and  the  number  of  revolutions 
per  minute  of  each  is  given: 

Rule  3.  —  The  diameter  of  the  driving  pulley  equals  the 
product  of  the  diameter  and  the  number  of  revolutions  of 
the  driven  pulley  divided  by  the  number  of  revolutions  of  the 
driving  pulley. 

0,  ,  .      *=£  .  | 

EXAMPLE.  —  The  driving  pulley  makes  100  revolutions  per  minute, 
the  driven  pulley  makes  75  revolutions  per  minute  and  is  18  inches  in 
diameter;  what  is  the  diameter  of  the  driving  pulley  ? 

SOLUTION.  —  Applying  the  rule  just  given  and  substituting,  we  have 
D  =  ^-  =  134  in.  Ans. 

16.  The  diameter  and  number  of  revolutions  per  minute 
of  the  driving  pulley  being  given,  to  find  the  diameter  of  the 
driven  pulley,  which  must  make  a  given  number  of  revolu- 
tions per  minute  : 

Rule  4.  —  The  diameter  of  the  driven  pulley  equals  the  prod- 
uct of  the  diameter  and  the  number  of  revolutions  of  the 
driving  pulley  divided  by  the  number  of  revolutions  of  the 
driven  pulley. 

,      DN 
Or,  d  =  -  . 


EXAMPLE.—  The  diameter  of  the  driving  pulley  is  13|  inches  and  it 
makes  100  revolutions  per  minute;  what  must  be  the  diameter  of  the 
driven  pulley  to  make  75  revolutions  per  minute  ? 

SOLUTION.—  Applying  rule  4  we  have  d  =  —  2_2l  -  =  jg  in.    Ans. 

17.  To  find  the  number  of  revolutions  per  minute  of  the 
driven  pulley,  its  diameter  and  the  diameter  and  the  number 
of  revolutions  per  minute  of  the  driving  pulley  being  given: 


12  MACHINE  ELEMENTS.  §  5 

Rule  5. —  The  number  of  revolutions  of  the  driven  pulley 
equals  the  product  of  the  diameter  and  the  number  of  revolu- 
tions of  the  driver  divided  by  the  diameter  of  the  driven  pulley. 

DN 
Or,  n  =  -r. 

EXAMPLE. — The  driving  pulley  is  13|  inches  in  diameter  and  makes 
100  revolutions  per  minute;  how  many  revolutions  will  the  driven 
pulley  make  in  1  minute  if  it  is  18  inches  in  diameter  ? 

D  N 

SOLUTION.— Formula,  n  —  — y-. 

a 

Substituting,  we  have  n  =  13^  *  10°  _  75  R.  p.  M.     Ans. 
lo 

18.  To  find  the  number  of  revolutions  per  minute  df 
the  driving  pulley,  its  diameter  and  the  diameter  and  the 
number  of  revolutions  per  minute  of  the  driven  pulley  being 
given : 

Rule  6. —  The  number  of  revolutions  of  the  driving  pulley 
equals  the  product  of  the  diameter  and  the  number  of  revolu- 
tions of  the  driven  pulley  divided  by  the  diameter  of  the 
driving  p  u  I  ley. 

"*-% 

EXAMPLE. — The  driven  pulley  is  18  inches  in  diameter  and  makes 
75  revolutions  per  minute;  how  many  revolutions  will  the  driving 
pulley  make  in  1  minute  if  it  is  13£  inches  in  diameter  ? 

SOLUTION.— Formula,  TV-—  -^. 

Substituting,  we  have  N=  18  *  75  =  100  R.  P.  M.     Ans. 


BELTS. 

19.  A  belt  is  a  flexible  connecting  band  that  drives  a 
pulley  by  its  frictional  resistance  to  slipping  at  the  surface 
of  the  pulley.  Belts  are  most  commonly  made  of  leather, 
cotton,  or  rubber,  and  are  united  in  long  lengths  by  cement- 
ing, riveting,  or  lacing. 


§  5  MACHINE  ELEMENTS.  13 

20.  Leather  belts   are   made  single   and   double.      A 
single  belt  is  one  composed  of  a  single  thickness  of  leather; 
a  double  belt  is  one  composed  of  two  thicknesses  of  leather 
cemented  and  riveted  together  the  whole  length  of  the  belt. 

21.  Cotton  belts  are  in  use  to  some  extent,  as  are  also 
belts  made  of  a  number  of  layers  of  duck  sewed  together  and 
impregnated  with  a  preparation  rendering  them  waterproof. 
These  belts,  in  accordance  with  the  number  of  layers  or 
"plys, "  are  called  two-ply,  three-ply,  etc.     Four-ply  cotton 
and  duck  belting  is  about  equal  to  single  leather  belting, 
and  eight-ply  to  double  leather  belting. 

22.  Rubber  belts  are  especially  adapted  for  use  in  damp 
or  wet  places ;  they  will  endure  a  great  degree  of  heat  or 
cold  without  injury,   are  quite  durable,  and  are  claimed  to 
be  less  liable  to  slip  than  leather  belts. 


CALCULATIONS  FOR  BELTS. 

23.  To  Find  the  Length  of  a  Belt. — In  practice,  the 
necessary  length  for  a  belt  to  pass  around  pulleys  that  are 
already  in  their  position  on  a  shaft  is  usually  obtained  by 
passing  a  tape  line  around  the  pulleys,  the  stretch  of  the  tape 
line  being  allowed  as  that  necessary  for  the  belt.  The  lengths 
of  open-running  belts  for  pulleys  not  in  position  can  be 
obtained  approximately  as  follows : 

Rule  7. —  The  length  of  a  belt  for  open-running  pulleys 
equals  3^  times  one-half  the  sum  of  the  diameters  of  the 
pulleys  plus  2  times  the  distance  between  the  centers  of  the 
shaft. 

Let       D  •=•  diameter  of  one  pulley  in  inches; 

d  =  diameter  of  the  other  pulley  in  inches; 
L  =  distance  between  the  centers  of  the  shafts 

in  inches; 
B  —  length  of  the  belt  in  inches. 

Then,  B  =  \ 


14  MACHINE  ELEMENTS.  §  5 

EXAMPLE.— The  distance  between  the  centers  of  two  shafts  is 
9  feet  7  inches ;  the  diameter  of  the  large  pulley  is  36  inches  and  the 
diameter  of  the  small  one  is  14  inches ;  what  is  the  necessary  length  of 
the  belt  ? 

SOLUTION. — Applying  the  rule  just  given  and  substituting  the 
values  given,  we  have,  since  9  feet  7  inches  =  115  inches, 

B  -  8j  (36  +  14)  +  2  X  115  =  Slli  in.,  or  25  ft.  llj  in.     Ans. 

The  length  of  crossed  belts  cannot  be  determined  by  any 
simple  calculation,  it  being  a  rather  difficult  mathematical 
problem. 

24.  To  Find  the  Width  of  Belts. — A  belt  should  be 
wide  enough  to  bear  safely  and  for  a  reasonable  length  of 
time  the  greatest  tension  that  will  be   put  upon  it.     This 
will  be  the  tension  of  the  driving  side.     The  safe  tension  for 
single  belts  may  be  taken  as  60  pounds  per  inch  of  width; 
single  belts  average  T3¥  inch  in  thickness.     The  tension  on 
the  driving  side,  however,  does  not  represent  the  force  tend- 
ing  to  turn  the   pulley.     The    force   tending    to  turn  the 
pulley,  or  the  effective  pull,  is  the  difference  in  tension 
between  the  driving  side  and  the  slack  side  of  the  belt.     The 
tension  on  the  driving  side  depends  on  three  factors:  the 
effective  pull  of  the  belt,   the  coefficient  of  friction  between 
the  belt  and  pulley,  and  the  size  of  the  arc  of  contact  of  the 
belt  on  the  smaller  pulley. 

25.  The  effective  pull  that  may  be  allowed  per  inch  of 
width  for  single  leather  belts  with  different  arcs  of  contact 
(the  arc  in  which  the  belt  touches  the  smaller  pulley),  is 
given  in  Table  I. 

26.  To   Find   the   Arc    of  Contact. — The  arc  of  con- 
tact in  degrees,  or  as  a   fraction  of  the  circumference,  can 
be  determined,  practically,  as  follows:  Stretch  a  string  over 
the  two  pulleys  to  represent  the  belt.     Then,  take  another 
string,  wrap  it  around  the  small  pulley  and  cut  it  off  so  that 
the  ends  meet.     This  represents  the  circumference  of  the 
small  pulley.     Now  take  a  third  string,  hold  one  end  at  the 


MACHINE  ELEMENTS. 
TABLE    I. 


18 


ALLOWABLE  BELT  PTJI/LS. 


Arc  Covered  by  Belt. 

Allowable  Effective  Pull 

Degrees. 

Fraction  of  Cir- 
cumference. 

Per  Inch  of  Width 
in  Pounds. 

90 

.250 

23.0 

112£ 

.312 

27.4 

120 

.333 

28.8 

135 

.375 

31.3 

150 

.417 

33.8 

157£ 

.437 

34.9 

180  or  over 

.500 

38.1 

beginning  of  the  arc  of  contact,  as  shown  by  the  string 
stretched  around  both  pulleys,  wrap  it  around  the  smaller 
pulley,  and  cut  it  off  at  the  end  of  the  arc  of  contact.  The 
length  of  this  last  string  represents  the  length  of  the  arc 
of  contact.  We  now  have  the  proportion:  the  length 
of  the  string  representing  the  circumference  :  the  length  of 
the  string  representing  the  arc  of  contact  ::  360  (the  number 
of  degrees  in  a  circle)  :  the  number  of  degrees  in  the  arc  of  con- 
tact. Whence,  the  number  of  degrees  in  the  arc  of  contact 
equals  the  quotient  obtained  by  dividing  the  product  of  the 
length  of  the  arc  of  contact  and  360  by  the  circumference 
of  the  pulley. 

To  obtain  the  fraction  of  the  circumference,  divide  the 
length  of  the  string  representing  the  arc  of  contact  by  the 
circumference. 

27.  To  use  the  table  for  finding  the  width  of  a  single 
leather  belt  for  transmitting  a  given  number  of  horsepower, 
we  have  the  following  rule, 


16  MACHINE  ELEMENTS.  §  5 

where  C  =  allowable  effective  pull,  from  table ; 
H  =  horsepower  to  be  transmitted; 
W=  width  of  single  belt  in  inches; 
V  =  velocity  of  belt  in  feet  per  minute. 

Rule  8. — Multiply  the  horsepower  to  be  transmitted  by 
S3, 000,  and  divide  this  product  by  the  product  of  the  velocity 
of  the  belt  and  the  allowable  effective  pull,  as  taken  from  the 
table.  The  quotient  will  be  the  width  of  the  belt. 

33, 000  # 


Or, 


vc 


EXAMPLE. — What  width  of  single  belt  is  needed  to  transmit  20  horse- 
power, the  arc  of  contact  on  the  small  pulley  being  135°  and  the  speed 
of  the  belt  1,500  feet  per  minute  ? 

SOLUTION.— According  to  Table  I,  the  allowable  effective  pull  for 
135°  is  31.3  pounds.  Then,  applying  rule  8,  we  have 

33,000  X  20 


28.  To  Find  the  Horsepower  of  a  Belt. — The  horse- 
power that  a  single  belt  will  transmit  is  given  by  the  follow- 
ing rule: 

Rule  9. — Multiply  together  the  effective  pull  taken  from 
the  table,  the  widtli  of  the  belt  in  inches,  and  the  speed  of  the 
belt  in  feet  per  minute.  Divide  the  product  by  33,000. 

CWV 


Or, 


33,000' 


29.  Speed  of  Belts. — By  applying  rule  8  to  the  same 
belt  running  at  different  velocities,  it  will  be  seen  that  the 
higher  the  velocity,  the  greater  is  the  horsepower  that  the 
same  belt  can  transmit,  and  from  rule  9  it  will  be  seen  that 
the  higher  the  speed  of  the  belt,  the  less  may  be  its  width 
to  transmit  a  given  horsepower.  From  this  it  follows  that 
a  belt  should  be  run  at  as  high  a  velocity  as  conditions  will 
permit,  the  maximum  velocity  allowable  for  a  laced  belt 
being  about  3,500  feet  per  minute  for  ordinary  single 
leather  and  double  leather  belts.  For  belts  spliced  by 


§  5  MACHINE  ELEMENTS.  17 

cementing,  where  the  splice  is  practically  as  strong  as  the  belt 
itself,  the  velocity  may  be  as  high  as  5,000  feet  per  minute. 
Cases  are  on  record  where  wide  main  belts  have  been  run  at 
as  high  a  velocity  as  6,000  feet  per  minute. 

3O.  In  choosing  a  proper  belt  speed,  due  regard  must 
be  paid  to  commercial  conditions.  While  a  high  speed  of 
the  belt  means  a  narrow  and,  consequently,  a  cheaper  belt, 
the  increased  cost  of  the  larger  pulleys  that  may  be  required 
may  offset  the  gain  due  to  the  high  belt  speed,  at  least  as 
far  as  first  cost  is  concerned. 

For  illustration,  let  the  problem  be  to  transmit  10  horse- 
power from  one  shaft  to  another.  Let  the  revolutions  of 
both  the  driven  and  the  driving  shaft  be  equal,  and  let  the 
shafts  make  200  revolutions  per  minute.  Choosing  a  belt 
speed  of  2,000  feet  per  minute,  the  width  of  a  single  belt 
to  transmit  the  given  horsepower  will  be,  by  rule  8,  say, 
4£  inches.  The  diameter  of  the  pulley  that  at  200  revolu- 
tions per  minute  will  give  a  belt  speed  of  2,000  feet  per 

C\    AAA    x/    "I  O 

minute  is  -  —  =  38J  inches,  nearly.      Taking  the 

&\J\J    /\     O.  - 


price  of  a  38-inch  cast-iron  pulley,  5-inch  face,  as  $15,  we 
have  the  price  of  two  pulleys  as  $30.  Let  the  distance 
between  the  pulleys  be  20  feet.  Then,  the  length  of  belt, 
according  to  rule  7,  is  50  feet  4  inches.  Taking  51  feet  as 
the  length  of  the  belt,  and  the  price  of  a  single  leather  belt 
4£  inches  wide  at  50  cents  per  foot,  the  price  of  the  belt  will 
be  $25.50.  Then,  the  cost  of  belt  and  pulleys,  not  count- 
ing freight  or  express  charges,  etc.,  will  be  $25.50  -f-  $30 
—  $55.50. 

Choosing  a  belt  speed  of  3,500  feet  per  minute,  the  width 
of  belt  will  be  2£  inches,  nearly.     The  proper  diameter  of 

Q  f)f\0  v  1  9 

the  pulley  is  ^       0  *  ,g-  =  66£  inches,  say,  67  inches.    The 
ZOO  X  o.l41o 

length  of  the  belt  will  be  59  feet,  about.  The  price  of  the 
belt  at  30  cents  per  foot  is  $17.70.  The  price  of  two 
67-inch  pulleys  3^-inch  face  is,  say,  $80.  Then,  the  total 
first  cost  is  $80  +  $17.70  =  $97.70,  showing  that  in  this 


18  MACHINE  ELEMENTS.  §  5 

particular  case  the  use  of  a  low  belt  speed  reduces  the  first 
cost  by  $97.70  —  $55.50  =  $42.20. 

The  above  illustration  is  not  intended,  and  must  not  "be 
construed,  to  be  an  argument  against  high  belt  speed;  it 
simply  shows  the  advisability  of  considering  the  commercial 
features  in  each  and  every  case.  In  many  cases  it  will  be 
found  that  the  narrow,  high-speed  belt  is  by  far  the  more 
economical  one  to  use.  .  , 

31.  Double  belts  are  made  of  two  single  belts  cemented 
and,  usually,  riveted  together  their  whole  length,  and  are 
used  where  much  power  is  to  be  transmitted.  As  the 
effective  pull  for  single  belts,  as  given  in  Table  I,  is  based 
primarily  on  the  strength  through  the  lace  holes,  a  double 
belt,  which  is  twice  as  thick,  should  be  able  to  transmit 
twice  as  much  power  as  a  single  belt,  and  in  fact  more  than 
this,  where,  as  is  quite  common,  the  ends  of  the  belt  are 
cemented  instead  of  laced. 

Where  double  belts  are  used  on  small  pulleys,  however, 
the  contact  with  the  pulley  face  is  less  perfect  than  it  would 
be  if  a  single  belt  were  used,  owing  to  the  greater  rigidity  of 
the  former.  More  work  is,  also,  required  to  bend  the  belt  as 
it  runs  over  the  pulley  than  in  the  case  of  the  more  pliable 
single  belt,  and  the  centrifugal  force  tending  to  throw  the 
belt  from  the  pulley  also  increases  with  the  thickness. 
Moreover,  in  practice,  it  is  seldom  that  a  double  belt  is  put 
on  with  twice  the  tension  of  a  single  belt.  For  these 
reasons,  the  width  of  a  double  belt  required  to  transmit  a 
given  horsepower  is  generally  assumed  to  be  seven-tenths 
the  width  of  a  single  belt  to  transmit  the  same  power.  On 
this  basis,  rules  8  and  9  become  for  double  belts,  by  multi- 
plying rule  8  by  T7T  and  dividing  rule  9  by  T\,  as  follows: 

Rule  1O. —  To  find  the  widtJi  of  a  double  belt,  multiply  the 
horsepower  to  be  transmitted  by  23, 100.  Divide  this  product 
by  the  product  of  the  velocity  of  the  belt  and  the  allowable 
effective  pull,  as  taken  from  the  table. 


u/     23, 100  H 
Or,  W= — .^     . 


§  5  MACHINE  ELEMENTS.  19 

Rule  11.  —  To  find  the  horsepower  that  a  double  belt  can 
transmit,  multiply  together  the  effective  pull  taken  from  the 
table,  the  width  of  the  belt,  and  its  velocity.  Divide  the 
product  by  23,  100. 

Or 


23,100' 

EXAMPLE  1.  —  What  width  of  double  belt  is  required  to  transmit 
20  horsepower,  the  arc  of  contact  on  the  smaller  pulley  being  135°  and 
the  speed  of  the  belt  1,500  feet  per  minute  ? 

SOLUTION.  —  According  to  Table  I,  the  effective  pull  is  31.3  pounds. 
Then,  by  rule  1O, 

23,100  x  20 
W=  r500  X  31.3  =  10  m"  nearly"     AnS" 

EXAMPLE  2.  —  What  horsepower  can  be  transmitted  by  a  6-inch 
double  belt  running  at  400  feet  per  minute  with  an  arc  of  contact  of  180°  ? 

SOLUTION.  —  According  to  Table  I,  the  effective  pull  is  38.1  pounds. 
Applying  rule  11,  we  have 


THE  CARE  AND  USE  OP  BELTS. 

32.  Leather  Belts. — It  is  a  much  disputed  question  as 
to  which  side  of  the  belt  should  be  run  next  to  the  pulley. 
The  more  common  practice,  it  is  believed,  is  to  run  the  belt 
with  the  hair  or  grain  side  nearest  the  pulley.  This  side  is 
harder  and  more  liable  to  crack  than  the  flesh  side.  By 
running  it  on  the  inside  the  tendency  is  to  cramp  or  com- 
press it  as  it  passes  over  the  pulley,  while  if  it  ran  on  the 
outside,  the  tendency  would  be  for  it  to  stretch  and  crack. 
The  flesh  side  is  the  tougher  side,  but  for  the  reason  given 
above  the  life  of  the  belt  will  be  longer  if  the  wear  comes 
upon  the  grain  side.  The  lower  side  of  the  belt  should  be 
the  driving  side,  the  slack  side  running  from  the  top  of  the 
driving  pulley.  The  sag  of  the  belt  will  then  cause  it  to 
encompass  a  greater  length  of  the  circumference.  Long 
belts,  running  in  any  other  direction  than  vertical,  work 


20  MACHINE  ELEMENTS.  §  5 

better  than  short  ones,  as  their  weight  holds  them  more 
firmly  to  their  work. 

It  is  bad  practice  to  use  rosin  to  prevent  slipping.  It 
gums  the  belt,  causes  it  to  crack,  and  prevents  slipping  for 
only  a  short  time.  If  a  belt,  in  good  condition,  persists  in 
slipping,  a  wider  belt  should  be  used.  Sometimes  larger 
pulleys  on  the  driving  and  driven  shafts  are  of  advantage,  as 
they  increase  the  belt  speed  and  reduce  the  stress  on  the 
belt.  Belts  may  be  kept  soft  and  pliable  by  oiling  them  once 
a  month  with  castor  oil  or  neatsfoot  oil. 

33.  The  Flapping  of  Belts. — One  of  the  most  annoy- 
ing troubles  experienced  with  belting  of  all  kinds  is  the 
violent  flapping  of  the  slack  side.  Flapping  may  be  due  to 
any  one  of  several  causes,  or  to  a  combination  of  them. 
The  most  usual  cause  is  that  one  or  both  of  the  pulleys  run 
out  of  true.  The  belt  is  then  alternately  stretched  and 
released,  and  while  this  may  not  cause  flapping  at  one  speed, 
it  will  usually  do  so  at  a  higher  speed.  If  the  belt  is  rather 
slack,  tightening  it  somewhat  may  cure  or  alleviate  the  flap- 
ping. The  most  obvious  and  best  remedy,  but  the  most 
expensive,  is  to  turn  the  pulleys  to  run  true. 

Pulleys  being  out  of  line  with  each  other  are  another  pro- 
lific source  of  flapping,  especially  when  one  or  both  run  out 
of  true.  First,  bring  the  pulleys  in  line;  if  this  fails,  tighten 
the  belt  if  it  is  rather  loose.  If  no  improvement  is  noticed 
and  it  is  not  possible  to  turn  the  pulleys,  try  to  lower  the 
belt  speed  a  little,  either  by  the  substitution  of  smaller 
pulleys  or  by  changing  the  speed  of  the  driven  shaft,  accord- 
ing to  circumstances. 

With  belts  running  at  speeds  above  4,000  feet  per  minute, 
flapping  may  occur  when  the  pulleys  are  perfectly  true  and 
in  line  with  each  other,  even  when  the  belt  has  the  proper 
tension.  This  is  believed  to  be  due  to  air  becoming 
entrapped  between  the  face  of  the  pulley  and  the  belt.  At 
any  rate,  it  has  been  observed  that  perforating  the  belt  with 
a  series  of  small  holes  will  cure  this  trouble.  Perforated 
belts  may  now  be  bought  in  the  market. 


§  5  MACHINE  ELEMENTS.  21 

Lack  of  steadiness  in  running,  due  either  to  sudden  varia- 
tions in  the  speed  of  the  engine,  or  sudden  changes  in  the 
load  of  the  machines  driven  by  the  belt,  will  produce  a  flap- 
ping that  it  is  almost  impossible  to  cure.  The  only  known 
cure  is  to  take  such  steps  as  will  insure  steady  running,  as 
for  instance,  increasing  the  weight  of  the  flywheel  on  the 
engine,  or  placing  a  flywheel  instead  of  a  pulley  on  the  driven 
machine. 

The  belt  not  being  joined  square  will  also  cause  flapping, 
especially  when  the  belt  is  running  at  a  rather  high  speed. 
The  remedy  is  to  unlace  or  unfasten  the  joint  and  make  it 
square. 

Too  great  a  distance  between  the  pulleys  may  also  cause 
flapping.  In  general,  the  distance  between  the  pulleys 
should  not  exceed  15  feet  for  belts  up  to  4  inches  in  width ; 
20  feet  for  belts  above  4  and  below  12  inches;  25  feet  for 
belts  above  12  inches  and  below  18  inches  ;  and  30  feet  for 
larger  belts.  The  distances  here  given  are  occasionally 
exceeded  considerably,  but  as  no  experiments  have  ever 
been  made  public  that  would  enable  a  fairly  correct  formula 
to  be  deduced  for  the  distance  between  pulleys  when  the 
belt  speed,  width  of  belt,  and  effective  pull  are  known,  they 
must  be  taken  as  representing  average  practice. 

A  horizontal  belt  is,  by  many,  considered  to  have  the 
proper  tension  when  it  has  about  1  inch  of  sag,  while  in 
motion,  for  every  8  feet  between  the  pulleys.  For  belts 
other  than  horizontal,  this  should  be  less,  there  being  no 
sag  at  all  for  vertical  belts. 


JOINING  THE  ENDS  OF  BELTS. 

34.  Lacing. — The  ends  of  a  belt  may  be  joined  by 
lacing,  sewing,  riveting,  or  cementing.  Many  ways  of  lacing 
belts  are  used.  A  very  satisfactory  method  for  belts  up  to 
3  inches  in  width  is'  shown  in  Fig.  16.  Cut  the  ends  of  the 
belt  square,  using  a  sharp  knife  and  a  try  square.  Punch  a 
row  of  holes  according  to  the  width  of  the  belt,  punching 


22 


MACHINE  ELEMENTS. 


f 


-i 


corresponding  holes  exactly  opposite  each  other  in  each  end 
of  the  belt,  using  3  holes  in  belts  up  to  2  inches  wide,  and 
5  holes  in  belts  between  2  and  3  inches  wide.     The  number 
of  holes  in  the  row  should  always 
be  uneven  for  the  style  of  lacing 
shown.     In  the  figure,  A  is  the 
outside  of   the   belt   and  B  the 
side  running  nearest  the  pulley. 
The  lacing  should  be  drawn  half- 
way through  one  of  the  middle 
holes   from   the    under   side,    as 
at  1-  before  going  any  further, 
it  is  well  to  see  to  it  that  the  belt 
FIG-  16-  has  no  twists  in  it,  or,  in  the  case 

of  a  crossed  belt,  that  it  has  not  been  given  a  wrong  twist. 
The  same  side  of  the  belt  should  run  over  both  pulleys, 
which  will  be  the  case  with  a  crossed  belt  if  it  has  been 
twisted  correctly.  Having  made  sure  that  the  belt  is  fair, 
pass  the  end  of  the  lace  on  the  upper  side  of  the  belt 
through  2  under  the  belt  and  up  through  3,  back  again 
through  2  and  3,  through  4  and  up  through  5,  where  an 
incision  is  made  in  one  side  of  the  lacing,  which  forms  a 
barb  that  will  prevent  the  end  from  pulling  through.  Lace 
the  right-hand  side  in  the  same  manner.  The  lacing  may 
advantageously  be  carried  on  at  once  to  the  right  and  left 
alternately. 

35.  For  belts  wider  than  3  inches,  the  lacing  shown  in 
Fig.  17  is  a  good  one.  As  will  be  observed,  there  are  two 
rows  of  holes.  The  number  of  holes  in  the  row  nearest  the 
joint  should  exceed  by  one  the  number  of  holes  in  the 
second  row.  For  belts  up  to  4£  inches  wide,  use  3  holes  in 
the  row  nearest  the  joint  and  2  holes  in  the  second  row. 
For  belts  up  to  6  inches  wide,  use  4  and  3  holes,  respect- 
ively. For  larger  belts,  make  the  total  number  of  holes  in 
each  end  either  one  or  two  more  than  the  number  of  inches 
of  width,  with  the  object  of  getting  an  odd  total  number  of 
holes.  For  example,  for  a  10-inch  belt,  the  total  number  of 


MACHINE  ELEMENTS. 


23 


holes  should  be  10  +  1  —  11.  For  a  13-inch  belt,  it  should 
be  13  +  2  =  15  holes.  The  outside  holes  of  the  first  row 
should  not  be  nearer  the  edges  of  the  belt  than  f  inch,  nor 
should  the  first  row  be  nearer  the  joint  than  £•  inch.  The 
second  row  should  be  at  least 
If  inches  from  the  end.  In 
the  figure,  A  is  the  outside 
and  B  the  side  nearest  the 
pulley.  Begin  at  one  of  the 
center  holes  in  the  outside  row, 
as  1,  and  continue  through  #, 
S,4,S,  6,  7,  4,3,  8,3,  etc. 

Another  method  is  to  begin 
the  lacing  on  one  side  instead 
of  in  the  middle.  This  method 
will  give  the  rows  of  lacing  on 
the  under  side  of  the  belt  the 
same  thickness  all  the  way  across.  The  lacing  should  not  be 
crossed  on  the  side  of  the  belt  that  runs  next  to  the  pulley. 

Lacing  affords  convenient  means  of  shortening  belts  when 
they  stretch  and  of  increasing  their  tension.  If  the  belts 
are  at  all  large,  the  ends  of  the  belt  should  be  drawn  together 
by  belt  clamps. 

36.  Cementing. — Cementing  makes  probably  the  best 
kind  of  a  joint  ever  devised.  It  has  the  serious  disadvantage, 
however,  that  the  stretch  of  the  belt  cannot  be  readily  taken 
up,  and,  hence,  tightening  pulleys  must  be  used  when  the 
center-to-center  distance  of  the  pulleys  is  not  adjustable. 


In  dynamo  driving,  where  endless  belts  are  used  to  the 
exclusion  of  all  others,  the  dynamo  is  usually  mounted  on  a 
slide,  so  that  the  tension  of  the  belt  can  be  adjusted.  For 
a  cemented  joint,  the  ends  of  the  belt  should  be  pared  down 
with  a  very  sharp  knife  to  the  form  shown  in  Fig.  18,  which 
shows  a  form  that  is  recommended  by  belt  manufacturers. 


24  MACHINE  ELEMENTS.  §  5 

Warm  the  belt  ends  near  a  fire,  apply  the  belt  cement  while 
hot,  and  press  the  joint  together  by  two  boards,  one  on  the 
top  and  one  on  the  bottom.  Belt  cement  can  be  obtained 
of  any  dealer  in  engineer's  supplies  and  full  directions  for 
using  it  will  always  be  found  on  the  can.  These  directions 
should  be  implicitly  followed. 

If  belt  cement  cannot  be  obtained,  a  good  cement  may  be 
made  by  melting  together  over  a  slow  fire  16  parts  of  gutta 
percha,  4  parts  of  india  rubber,  2  parts  of  pitch,  1  part  of 
shellac,  and  2  parts  of  linseed  oil,  by  weight.  Cut  all  ingre- 
dients very  small,  mix  well,  and  use  while  hot. 

37.  Stretch  of  Belts. — New  leather  belts  will  stretch 
from  one-fourth  to  one-half  of  an  inch  per  foot  of  length, 
and  hence  must  be  taken  up  until  the  limit  of  stretch  has 
been  reached.     Rubber  belts  are  said  to  stretch  continuously. 
Cotton  and  duck  belts  are  said  not  to  stretch  with  use. 

38.  Precautions  to  l>e  Observed  When  Using  Rubber 
Belts. — When  rubber  belts  are  used,  animal  oils  or  animal 
grease  should  never  be  used   on  them.     If  the  belt  should 
slip,  it  should  be  lightly  moistened  on  the  side  nearest  the 
pulley  with  boiled  linseed  oil. 


EXAMPLES   FOR   PRACTICE. 

1.  The  main  line  shaft  is  driven  at  90  revolutions  per  minute  and 
is  to  drive  another  shaft  at  120  revolutions  per  minute;  the  latter  shaft 
carries  a  pulley  20  inches  in  diameter.     How  large  a  pulley  should  be 
used  on  the  main  line  shaft  ?  Ans.  26J  in. 

2.  The  belt  wheel  of  an  engine  is  10  feet  in  diameter  and  makes 
65  revolutions  per  minute ;  the  line  shaft  is  to  run  at  150  revolutions 
per  minute.     What  size  pulley  should  be  used  ?  Ans.  52  in. 

3.  The  driving  pulley  is  48  inches  in  diameter  and  makes  90  revolu- 
tions per  minute.     The  driven  pulley  being  20  inches  in  diameter,  how 
many  revolutions  per  minute  will  the  driven  shaft  make  ? 

Ans.  216  R.  P.  M. 

4.  The  belt  wheel  of  an  engine  being  5  feet  in  diameter,  the  driven 
pulley  45  inches,  and  the  driven  shaft  to  make  90  revolutions  per 
minute,  what  should  be  the  number  of  revolutions  of  the  belt  wheel  ? 

Ans.  67*  R.  P.  M. 


§  5  MACHINE  ELEMENTS.  25 

5.  Two  pulleys,  48  and  32  inches  in  diameter,   respectively,  are 
10  feet  6  inches  from  center  to  center.     What  is  the  length  of  an  open 
belt  for  these  pulleys  ?  Ans.  31  ft.  10  in. 

6.  A  belt  speed  of  2,400  feet  per  minute  having  been  chosen,  what 
width  of  single  belt  will  be  needed  to  transmit  30  horsepower  ?    The 
pulleys  over  which  the  belt  runs  are  equal  in  size.     Ans.  11  in.,  nearly. 

7.  What  horsepower  can  be  transmitted  by  a  single  belt  11  inches 
wide  traveling  at  the  rate  of  1,000  feet  per  minute  over  a  small  pulley 
on  which  its  arc  of  contact  is  150°  ?  Ans.  11±  H.  P.,  nearly. 

8.  With  a  belt  speed  of  1,300  feet  per  minute,  what  width  of  double 
belt  will  be  required  to  transmit  5  horsepower  over  pulleys  equal  in 
diameter?  Ans.  2£  in.,  nearly. 


WHEEL    WORK. 


DRIVER  AND  FOLLOWER. 

39.  A  combination  of  wheels  and  axles,  as  in  Fig.  19,  is 
called  a  train.  The  wheel  in  a  train  to  which  motion  is 
imparted  from  a  wheel  on  another  shaft,  by  such  means  as  a 
belt  or  gearing,  is  called  the  driven  wheel  or  follower; 
the  wheel  that  imparts  the  motion  is  called  the  driver. 

It  will  be  seen  that  the  wheel  and  axle  bears  the  same 
relation  to  the  train  that  a  simple  lever  does  to  the  com- 
pound lever.  Letting  Dv,  D^  D3,  etc.  represent  the  diam- 
eters of  the  driven  wheels,  and  d^  d^  d^  etc.  the  diameters 
of  the  different  drivers,  we  have  the  following  rule : 

Rule  13. —  The  continued  product  of  the  force  and  the 
diameters  of  the  driven  wheels  equals  the  continued  product 
of  the  weiglit,  tJie  diameter  of  the  drum  that  moves  the  weight, 
and  the  diameters  of  the  drivers. 

Or,  Px  Dl  X  A  X  Dv  etc.  =  IVx  <  X  <  X  </„  etc. 

Since  the  radii  of  several  circles  are  in  the  same  propor- 
tion to  one  another  as  the  diameters  of  the  same  circles,  it 
follows  that  the  radii  may  be  used  instead  of  the  diameters 
in  the  above  rule. 


26  MACHINE  ELEMENTS.  §  5 

EXAMPLE.— If  the  radius  of  the  pulley  A,  Fig.  19,  is  20  inches,  of  C 
15  inches,  and  of  E  24  inches,  and  the  radius  of  the  drum  F  is  4  inches, 
of  the  pinion  D  5  inches,  and  of  the  pinion  B  4  inches,  how  great 
a  weight  will  a  force  of  1  pound  applied  at  P  raise  ? 


® 


FIG.  19. 
SOLUTION. — Applying  rule  12,  we  have 

1X20X15X24=  J^  X  4  X  5  X  4, 

n  90fi 
or  W=.^p  =  901b.     Ans. 

oU 

4O.  Although  the  combination  of  wheels  in  this  example 
enables  the  lifting  of  a  weight  90  times  as  great  as  the  force 
applied,  it  has  been  necessary  to  exert  the  force  through  a 
distance  90  times  the  height  through  which  the  weight  was 
raised.  It  is  a  universal  law  in  the  application  of  machines 
that  2vhenever  there  is  a  gain  in  power  without  a  correspond- 
ing increase  in  the  initial  force,  there  is  a  loss  in  speed ;  this 
is  true  of  any  machine. 


MACHINE  ELEMENTS. 


•27 


In  the  example,  if  P  were  to  move  the  entire  90  inches  in 
1  second,  W  would  move  only  1  inch  in  the  same  period  of 
time. 

41.  Instead  of  using  the  diameter  or  radius  of  a  gear,  as 
in  the  last  example,  the  number  of  teeth  may  be  used  when 
computing  the  weight  that  can  be  raised,  or  the  velocity. 

EXAMPLE.— The  radius  of  the  pulley  A,  Fig.  19,  is  40  inches  and 
that  of  F  is  12  inches.  The  number  of  teeth  in  B  is  9,  in  C  27,  in 
D  12,  and  in  E  36.  If  the  weight  to  be  lifted  is  1,800  pounds,  how 
great  a  force  at  P  is  it  necessary  to  apply  to  the  belt  ? 

SOLUTION. — Let  P  represent  the  force;  then,  by  the  rule 

P  X  40  X  27  X  36  =  1,800  X  12  X  9  X  12, 
or  P  X  38,880  =  2,332,800. 

Hence, 


38,880 
=  the  amount  of  force  necessary  to  apply  to  the  belt.     Ans. 


GEAR- WHEELS. 

42.  A  wheel  that  is  provided  with  teeth  that  mesh  with 
similar  teeth  on  another  wheel  is  called  a  gear-wheel,  or 
gear.  In  Fig.  20  is  shown  a  spur  gear.  On  spur  gears 


FIG.  20. 


28  MACHINE  ELEMENTS.  §  5 

the  teeth  are  usually  parallel  to  the  axis  of  the  wheel  or  to, 
its  shaft. 

Gears  are  said  to  be  in  mesh  when  the  teeth  of  two  wheels, 
respectively,  engage  each  other  or  interlock. 

43.  In  Fig.  21  is  shown  a  pair  of  bevel  gears  in  mesh. 
Of  the  two  wheels  shown,  one  is  smaller  than  the  other; 
when  both  wheels  of  a  pair  of  bevel  gears  are  of  the  same 
diameter  they  are  called  miter  gears.  In  Fig.  22  is  shown 


a  pair  of  miter  gears  in  mesh.  It  is  obvious  that  the  angle 
that  the  teeth  of  these  gears  make  with  the  axis  of  the  shaft 
must  be  45°. 

44.     Of  a  pair  of  gear-wheels  (either  spur  or  bevel)  having 
different  diameters,  the  smaller  is  called  a  pinion. 


§  5  MACHINE  ELEMENTS.  29 

In  Fig.  23  is  shown  a  revolving  screw,  or  worm,  as  it  is 
called,  that  meshes  with  a  worm-wheel.  It  is  used  to 
transmit  motion  from  one  shaft 
to  another  at  right  angles  to  it. 

As  the  worm  is  nothing  else 
than  a  screw,  each  revolution 
given  to  it  will  rotate  the 
wheel  a  distance  equal  to  the 
pitch  of  the  worm ;  consequent- 
ly, if  there  are  40  teeth  in  the 
worm-wheel,  a  single-threaded 
worm  must  make  40  revolu- 
tions in  order  to  turn  the  wheel 
once. 

45.  A  rack  is  a  straight 
bar  that  has  gear-teeth  cut  on 
it.  It  may  be  considered  as  part  FIG.  23. 

of  a  gear-wheel  in  which  the  diameter  is  infinitely  large.  The 
teeth  of  racks  are  proportioned  by  the  same  rules  as  those  of 
gear-wheels. 


TEETH  OF  GEAR-WHEELS. 

46.  The  object  in  designing  the  teeth  of  gear-wheels 
should  be  to  so  shape  them  that  the  motion  transmitted 
will  be  exactly  the  same  as  with  a  corresponding  pair  of 
wheels,  or  cylinders,  without  teeth  and  running  in  contact 
without  slipping.  Such  cylinders  are  called  pitch  cylinders, 
and  are  always  represented  on  the  drawing  of  a  gear-wheel 
by  a  line  called  the  pitch  circle  (see  Fig.  24).  The  pitch 
circle  is  also  called  the  pitch  line. 

The  diameter  of  the  pitch  circle  is  called  the  pitch 
diameter.  When  the  word  "diameter"  is  applied  to 
gears,  it  is  always  understood  to  mean  the  pitch  diameter, 
unless  especially  stated  as  the  "diameter  over  all"  or 
"diameter  at  the  root." 


30 


MACHINE  ELEMENTS. 


47.  Circular  and  Diametral  Pitch. — The  distance 
from  a  point  on  one  tooth  to  a  corresponding  point  on  the 
next  tooth,  measured  along  the  pitch  circle,  is  the  circular 


FIG.  24. 

pitch.  This  is  obtained  by  dividing  the  circumference 
(pitch  circle)  by  the  number  of  teeth,  and  is  used  in  laying 
out  the  teeth  of  large  gears  and,  also,  when  calculating 
their  strength. 

It  would  be  very  convenient  to  have  the  circular  pitch 
expressed  in  manageable  numbers  like  1-inch,  f  inch,  etc. ; 
but  as  the  circumference  of  a  gear  is  3.1416  times  its  diam- 
eter, this  requires  awkward  numbers  for  the  diameters. 
Thus,  a  wheel  of  40  teeth,  1-inch  pitch,  would  have  a  cir- 
cumference on  the  pitch  circle  of  40  inches  and  a  diameter 
of  12.732  inches.  Of  the  two  it  is  more  convenient,  in  the 
great  majority  of  cases,  to  have  the  diameters  expressed  in 
numbers  that  can  be  easily  handled.  In  order,  however,  to 


§  5  MACHINE  ELEMENTS.  31 

have  the  pitch  in  a  convenient  form  also,  a  pitch  has  been 
devised  that  is  expressed  in  terms  of  the  diameter  and  called 
the  diametral  pitch. 

The  diametral  pitch  is  not  a  measurement  like  the  circu- 
lar pitch,  but  a  ratio.  It  is  the  ratio  of  the  number  of  teeth 
in  the  gear  to  the  number  of  inches  in  the  diameter  ;  or,  it  is 
the  number  of  teeth  on  the  circumference  of  the  gear  for 
1  inch  diameter  of  the  pitch  circle.  It  is  obtained  by  divi- 
ding the  number  of  teeth  by  the  diameter. 

A  gear,  for  example,  has  60  teeth  and  is  10  inches  in 
diameter.  The  diametral  pitch  is  the  ratio  of  60  to 
10  =  \%  =  6  ;  this  gear  would  be  called  a  6-pitch  gear. 
From  the  definition  it  follows  that  teeth  of  any  particular 
diametral  pitch  are  of  the  same  size  and  have  the  same 
width  on  the  pitch  line,  whatever  may  be  the  diameter  of 
the  gear.  Thus,  if  a  12-inch  gear  has  48  teeth,  it  will  be 
4  pitch.  A  24-inch  gear  to  have  teeth  of  the  same  size  will 
have  twice  48,  or  96  teeth,  and  as  96  -H  24  =  4,  has  the  same 
diametral  pitch  as  before. 

48.  Other   Definitions. — The  other    necessary    defini- 
tions applying  to  the  parts  of  a  gear  can  be  readily  under- 
stood from  Fig.  24.     The  thickness  of  the  tooth  and  the 
width  of  the  space  are  measured  on  the  pitch  circle.     A 
tooth  is  composed  of  two  parts,   the  addendum,  or  part 
outside  of  the  pitch  circle,  and  the  root,  which  is  inside. 

A  line  through  the  outside  end  of  the  addendum  is  called 
the  addendum  circle,  or  addendum  line,  and  one  through 
the  inside  part  of  the  root  is  called  the  root  circle,  or 
root  line.  The  amount  by  which  the  width  of  the  space 
is  greater  than  the  thickness  of  the  tooth  is  called  the  back- 
lash, or  side  clearance. 

49.  Proportions  for  Gear-Teeth. — With  gears  of  large 
size,  and  often  with    cast   gears  of  all  sizes,   the  circular- 
pitch  system  is  used.      In  these  cases,  it  is  usual  to  have  the 
addendum,  whole  depth,  and  thickness  of  the  tooth  conform 
to  arbitrary  rules  based  on  the  circular  pitch. 


32  MACHINE  ELEMENTS.  §  5 

The  usual  proportions  are  for  cast  gears:  Addendum 
=  .3  X  circular  pitch.  Root  =  .4  X  circular  pitch.  Thick- 
ness of  tooth  =  .48  X  circular  pitch. 

The  gears  most  often  met  with  are  the  cut  gears  of  small 
and  medium  size  like  those,  for  example,  on  machine  tools, 
which  are  almost  invariably  diametral-pitch  gears.  The 
teeth  are  cut  from  the  solid  with  standard  milling  cutters, 
proportioned  with  the  diametral  pitch  as  a  basis.  This  sys- 
tem is  also  coming  into  use  for  cast  gearing.  In  all  dia- 
metral-pitch gears,  the  addendum,  in  inches,  is  made  equal 
to  1  divided  by  the  diametral  pitch,  and  the  working  depth 
to  twice  the  addendum.  The  end  clearance  is  usually  taken 
equal  to  ^  the  addendum  for  cut  gears,  though  The  Brown  & 
Sharpe  Manufacturing  Company  use  ^  the  thickness  of  the 
tooth  on  the  pitch  line  as  the  clearance.  The  side  clear- 
ance, or  ''backlash,"  is  barely  enough  to  give  a  good  work- 
ing fit,  and  seldom  exceeds  -5*5-  the  pitch. 

Using  the  above  proportions,  a  4-pitch  gear  will  have  the 
addendum  =  1  -4-  4  =  £  inch;  the  working  depth  will  be 
2  X  i=i  inch;  and  the  clearance,  if  made  £  the  addendum, 
|  X  i  =  -sV  incn-  Tne  whole  length  of  the  tooth  will  be 
£  +  ^  =  If  inch.  The  thickness  of  the  tooth  will  be  one- 
half  the  circular  pitch,  nearly.  In  a  10-pitch  wheel,  the 
addendum  will  be  -Jw  inch  and  the  length  of  the  tooth 
|£  inch;  in  a  2|-pitch,  it  will  be  1  H-  2£  —  f  inch  and  the 
length  ££  inch. 


FORMS  OF  GEAR-TEETH. 

50.  The  forms   of  teeth  used  in  ordinary  practice  form 
part  of  certain  curves  known  as  the  epicycloid,  Jiypocycloid, 
and  involute. 

51.  The  Eplcycloidal  Tooth. — In  the  so-called  epicy- 
cloidal  tooth,  which  more  properly  is  called  a  cycloidal  tooth, 
the  face  of  the  tooth  is  part  of  an  epicycloid,  and  the  flank, 
part  of  a  hypocycloid. 


MACHINE  ELEMENTS. 


33 


An  epicycloidal  curve  is  the  path  described  by  any  point 
of  a  circle  rolling,  without  slipping,  on  the  outside  of 
another  circle.  A  hypocycloid  is  the  path  described  by 
any  point  of  a  circle  rolling,  without  slipping,  on  the  inside 
of  another  circle. 

Epicycloidal  teeth  can  always  be  recognized  by  their 
appearance;  they  are  formed  by  two  curves  that,  com- 
mencing at  the  pitch  circle,  curve  in  opposite  directions. 
Fig.  25  clearly  exhibits  the  characteristic  tooth  form. 


PIG.  25. 

The  use  of  epicycloidal  teeth  is  being  gradually  aban- 
doned, as  they  possess  some  practical  defects,  the  chief 
defect  being  that  the  center-to-center  distance  of  the  two 
gears  must  be  practically  perfect  in  order  to  insure  a  uni- 
form velocity  of  the  driven  gear. 

52.  Involute  Teeth. — In  Fig.  26  is  shown  the  involute 
form  of  tooth,  which  is  composed  of  but  one  curve. 

The  involute  is  the  path  described  by  any  point  of  a 
string  that  is  being  wound  on  or  off  a  cylinder,  the  cylinder 
being  stationary.  In  the  involute  system,  the  sides  of  the 
teeth  of  the  rack  are  straight  lines,  as  shown  in  Fig.  26. 

Involute  teeth  have  two  great  advantages  over  epicycloi- 
dal teeth:  (1)  They  are  stronger  for  the  same  pitch,  as 
they  are  thicker  at  the  root.  (2)  The  gears  may  be  spread 


34  MACHINE  ELEMENTS.  §  5 

slightly  apart  so  that  their  pitch  circles  do  not  run  tangent  to 


FIG.  26. 


one  another,  without  affecting  the  perfect  action  of  the  teeth 
to  an  appreciable  extent. 


GEAR    CALCULATIONS. 

53.  The  Circular-Pitch  System. — For  calculating  the 
pitch  diameter,  number  of  teeth,  etc.    of  gear-wheels,  we 
have,  for  the  circular-pitch  system,  the  following  rules, 
where       P=  circular  pitch  in  inches; 

T  =  number  of  teeth ; 

D  =  pitch  diameter  of  the  gear  in  inches. 

54.  To  find  the  pitch  diameter  of  a  gear-wheel  in  inches, 
when  the  pitch  and  number  of  teeth  are  given: 

Rule    13. —  The  pitch  diameter  equals  the  product  of  the 
pitch  and  tlie  number  of  teeth  divided  by  3. 1416. 

PT 


Or, 


D  =  — — — 
3.1416' 


EXAMPLE. — What  is  the  diameter  of  the  pitch  circle  of  a  gear-wheel 
that  has  75  teeth  and  whose  pitch  is  1.675  inches  ? 
SOLUTION.— Applying  rule  13,  we  have 

=  40  in.     Ans. 


§  5  MACHINE  ELEMENTS.  35 

55.  To  find  the  number  of  teeth  in  a  gear-wheel  when 
the  diameter  and  pitch  are  given: 

Rule    14. — The  number   of  teeth   equals   the  product   of 
3. 1416  and  the  diameter  divided  by  the  pitch. 

„      3.1416  D 
Or,  T= -p . 

EXAMPLE. — The  diameter  of  a  gear-wheel  is  40  inches  and  the  pitch 
of  the  teeth  is  1.675  inches;  how  many  teeth  are  there  in  the  wheel  ? 

SOLUTION. — Applying  the  rule  just  given,  we  have 

T=  3.1416  X  40  =  75teeth      Ans 
l.b<5 

56.  To  find  the  pitch  of  a  gear-wheel  when  the  diameter 
and  the  number  of  teeth  are  given: 

Rule  15. —  The  pitch  of  the  teeth  equals  the  product  of 
3.  lJf.16  and  the  diameter  divided  by  the  number  of  teeth. 

3.1416 D 


EXAMPLE.—  The  diameter  of  a  gear-wheel  is  40  inches  and  it  has 
75  teeth  ;  what  is  the  pitch  of  the  teeth  ? 

SOLUTION.  —  By  rule  15,  we  have 

=  1.675  in.  pitch.     Ans. 


57.  The  Diametral-Pitch   System.  —  The  diameter  of 
the  gear-wheel,  the  number  of  teeth,  etc.  are  given  by  the 
following  rules,   where    Pd  =  diametral  pitch  ;  D0  =  outside 
diameter;  N  =  number  of  teeth;  and  the  other  letters  have 
the  same  meaning  as  in  the  three  preceding  rules. 

58.  To  find  the  pitch  diameter  of  the  gear-wheel  when 
the  number  of  teeth  and  the  pitch  are  given  : 

Kule  16.  —  Divide  the  number  .of  teeth  by  the  diametral 
pitch. 

i- 


36  MACHINE  ELEMENTS.  §  5 

EXAMPLE. — A  wheel  is  to  have  40  teeth,  4  pitch;  what  is  its  pitch 
diameter  ? 

SOLUTION.— By  applying  rule  16,  we  have 

40 
Z>=^  =  10in.     Ans. 

59.  To  find  the  diameter  over  all,  that  is,  the  diameter 
of  the  blank  from  which  the  gear-wheel   is  cut,  the  number 
of  teeth  and  the  diametral  pitch  being  given : 

Rule  17. — Add  2  to  the  number  of  teeth  and  divide  by  the 
dia  metral  pitch. 

A 

EXAMPLE. — In  the  last  example,  what  is  the  diameter  over  all  the 
blank? 

SOLUTION. — Applying  the  rule  just  given,  we  get 

=  10*  in.     Ans. 

60.  The  number  of  teeth  and  the  outside  diameter  of 
the  gear-wheel  being  known,  to  find  the  diametral  pitch: 

Rule  18. — Add  2  to  the  number  of  the  teeth  and  divide  by 
the  outside  diameter. 

p. 

EXAMPLE. — A  gear-wheel  has  60  teeth  and  is  6^  inches  in  diameter 
over  all ;  what  is  the  diametral  pitch  ? 

SOLUTION.— By  applying  rule  18,  we  get 

Pa  =  M£*  =  W,     Ans. 

61.  To  find  the  diametral  pitch,  the  number  of  teeth 
and  the  pitch  diameter  being  known: 

Rule  19. — Divide  the  number  of  teeth  by  the  pitch  diam- 
eter. 

**% 

EXAMPLE.— A  wheel  has  90  teeth  and  its  pitch  diameter  is  30  inches; 
what  is  the  diametral  pitch  ? 


§  5  MACHINE  ELEMENTS.  37 

SOLUTION. — By  rule  19,  we  get 

on 

Pd  =  g  =  3.     Ans. 

62.  The  pitch  diameter  and  the  diametral  pitch  being 
given,  to  obtain  the  number  of  teeth : 

Rule  2O. — Multiply  the  pitch  diameter  by  the  diametral 

pitch. 

Or,  N=DPd. 

EXAMPLE.— How  many  teeth  are  there  in  a  gear-wheel  having  a 
pitch  diameter  of  36  inches  and  a  diametral  pitch  of  5  ? 
SOLUTION. — Applying  the  rule  just  given,  we  get 
N=  36  X  5  =  180  teeth.     Ans. 

63.  The  diameter  over  all  and  the  diametral  pitch  being 
given,  to  find  the  number  of  teeth : 

Rule    21. — Subtract   2  from   the  product  of  the  outside 
diameter  and  the  diametral  pitch. 

Or,  N  =  D0  Pd  -  2. 

EXAMPLE. — How  many  10-pitch  teeth  has  a  gear-wheel  having  an 
outside  diameter  of  8^  inches  ? 
SOLUTION.. — By  rule  21,  we  get 

^=81x10-2  =  80166^1.     Ans. 

64.  The  standard  diametral  pitches  used  for  cut  gears 
are  as  follows:  2,  2£,  2|,  2f,  3,  3£,  4,  5,  6,  7,  8,  9,  10,  11,  12, 
14,  16,  18,  20,  22,  24,  26,  28,  30,  32,  36,  40,  48.     Gears  hav- 
ing a  diametral  pitch  differing  from  those  given  here  are, 
usually,  either  very  large  or  very  small. 

65.  Diameters   and   Distances   Between    Centers. — 

The  distance  between  the  centers  of  two  gear-wheels  being 

known  and  the  ratio  of  their  speeds,  the   diameter  of  the 

pitch  circle  of  the  smaller  wheel  is  given  by  the  following 

rule, 

where  A  =  center-to-center  distance; 

R  =  revolutions  per  minute  of  large  gear; 

r  =  revolutions  per  minute  of  small  gear;  , 

d  =  pitch  diameter  of  small  gear. 

//.  s.    /.— 16 


38  MACHINE  ELEMENTS.  §  5 

Rule  22. — Multiply  twice  the  center-to-center  distance  by 
the  number  of  revolutions  of  the  large  gear,  and  divide  the 
product  by  the  sum  of  the  revolutions  of  the  large  and  small 
gear. 

,      %AR 

Or,  d=R  +  ~f 

EXAMPLE. — Given,  the  distance  between  centers  =  5£  inches;  the 
small  gear  is  to  make  24  revolutions  for  every  8  revolutions  of  the 
large  gear.  What  is  the  diameter  of  each  gear  ? 

SOLUTION. — By  rule  22, 

<t=2^^<^  =  2f  in.,  the  diameter  of  the  small  wheel. 

Then,  as  the  center  distance  is  equal  to  the  sum  of  the  radii  of  the 

2f 
two  wheels;  the  radius  of  the  large  wheel  is  5|  —  -^  =  4|  in.,  and  its 

diameter  is  4|  X  2  =  8£  in.     Ans. 

66.  Speed  and  Number  of  Teeth.— To  calculate  the 
number  of  teeth  or  the  speed  of  one  of  two  gear-wheels  that 
are  to  gear  together : 

Let  N  —  number  of  revolutions  per  minute  of  the  driving 

wheel ; 
n  =  number  of  revolutions  per  minute  of  the  driven 

wheel; 

T  =  number  of  teeth  in  the  driving  wheel; 
t  =  number  of  teeth  in  the  driven  wheel. 

Rule  23. —  The  number  of  teeth  in  the  driving  wheel  equals 
the  product  of  the  number  of  teeth  and  number  of  revolutions 
of  the  driven  wheel  divided  by  the  number  of  revolutions  of 
the  driving  wheel. 

rr,        tn 

Or,  T=Jf. 

EXAMPLE.— The  driven  wheel  has  27  teeth  and  will  make  66  revolu- 
tions per  minute;  if  the  driving  wheel  makes  99  revolutions  per  min- 
ute, how  many  teeth  are  there  in  the  driving  wheel  ? 

SOLUTION. — Applying  rule  23,  we  have 


§  5  MACHINE  ELEMENTS.  39 

67.  The  number  of  revolutions  per  minute  of  the  driving 
wheel  and  the  driven  wheel  and  the  number  of  teeth  in  the 
driving  wheel  being  given,  to  find  the  number  of  teeth  in  the 
driven  wheel  : 

Rule  24.  —  The  number  of  teeth  in  the  driven  wheel  equals 
the  product  of  the  number  of  teeth  and  revolutions  per  minute 
of  the  driving  ^vheel  divided  by  the  number  of  revolutions  per 
minute  of  the  driven  wheel. 

Or,  , 

EXAMPLE.  —  The  driving  wheel  has  18  teeth  and  makes  99  revolu- 
tions per  minute,  and  the  driven  wheel  must  make  66  revolutions  per 
minute  ;  how  many  teeth  must  there  be  in  the  driven  wheel  ? 

SOLUTION.  —  Applying  the  rule  just  given,  we  have 


68.  The  number  of  teeth  in  the  driving  wheel  and  the 
driven  wheel  and  the  number  of  revolutions  per  minute  of 
the  driving  wheel  being  given,  to  find  the  number  of  revolu- 
tions per  minute  of  the  driven  wheel  : 

Rule  25.  —  The  number  of  revolutions  per  minute  of  the 
driven  wheel  equals  the  product  of  the  number  of  teeth  and 
number  of  revolutions  of  the  driving  wheel  divided  by  the 
number  of  teeth  of  the  driven  wheel. 

TN 
Or,  n  =  —  . 

EXAMPLE.  —  There  are  18  teeth  in  the  driving  wheel  and  it  makes 
99  revolutions  per  minute  ;  how  many  revolutions  per  minute  will  the 
driven  wheel  make  if  it  has  27  teeth  ? 

SOLUTION.  —  Applying  rule  25,  we  have 

n  =  ^—  °  =  66  R.  P.  M.     Ans. 

69.  The  number  of  teeth  in  the  driving  wheel  and  the 
driven  wheel  and  the  number  of  revolutions  per  minute  of 
the  driven  wheel  being  given,  to  find  the  number  of  revolu- 
tions per  minute  of  the  driving  wheel: 


40 


MACHINE  ELEMENTS. 


Rule  26. —  The  number  of  revolutions  of  the  driving  wheel 
equals  the  product  of  the  number  of  teeth  and  revolutions  of 
the  driven  ivheel  divided  by  the  number  of  teeth  of  the  driving 
wheel. 

#-'+ 

EXAMPLE  1.— If  there  are  27  teeth  in  the  driven  wheel  and  if  it 
makes  66  revolutions  per  minute,  how  many  revolutions  per  minute 
will  the  driving  wheel  make  if  it  has  18  teeth  ? 

SOLUTION. — Applying  the  rule  just  given,  we  have 


p 


Ans. 


EXAMPLE  2.— In  Fig.  27,  the  crank-shaft  makes  60  revolutions  per 
minute;  the  governor  pulley  is  4  inches  in  diameter;  the  bevel  gear  on 
the  governor  pulley  shaft  has  19  teeth;  the  bevel  gear  that  meshes 
with  it  and  drives  the  governor  has  30  teeth.  The  governor  is  to 
make  95  revolutions  per  minute;  what  should  be  the  size  of  the  pulley 
on  the  crank-shaft  ? 


12' 


FIG.  27. 

SOLUTION. — First  determine  the  number  of  revolutions  of  the  4-inch 
pulley  in  order  that  the  governor  shall  turn  95   times  per  minute. 
/  n      30  X  95 


Applying  rule 


Hi 


=  150  revolutions   of  gear   on 


pulley  shaft  =  revolutions  of  governor  pulley.     Now,  applying  rule  3, 
the  diameter  of  the  pulley  on  the  crank-shaft  = -^  =  —~ —  =  10  in. 

Ans. 


§  5  MACHINE  ELEMENTS.  41 

EXAMPLE  3. — In  Fig.  27,  the  flywheel  is  8  feet  in  diameter  and  drives 
a  5-foot  pulley  on  the  main  shaft.  A  14-inch  pulley  on  the  main  shaft 
drives  a  16-inch  pulley  on  the  countershaft.  A  12-inch  pulley  on  the 
countershaft  drives  a  12-inch  pulley  on  a  shaft  on  which  is  a  pinion 
that  meshes  into  a  large  gear  attached  to  the  face  plate  of  a  large  lathe, 
and  which  has  108  teeth.  How  many  teeth  must  the  pinion  have  in 
order  that  the  face  plate  may  make  9^  revolutions  per  minute  ? 

SOLUTION. — Applying  rule  5,  to  find  the  revolutions  per  minute  of 

8  X  60 
the  main  shaft,  — - —  =  96  R.  P.  M.     Applying  the  same  rule  again 

to   find   the  revolutions  of  the  countershaft,    — ^ —  =  84  R.  P.  M. 

16 

Applying  it  once  more  to  find  the  revolutions  of  the  pulley  that 
turns  the  small  gear,  12^84  =  84  R.  P.  M.  Applying  rule  33, 1Q8g*  9*" 
=  12  teeth  in  pinion  or  driver.  Ans. 


EXAMPLES.  FOR  PRACTICE. 

1.  The   driving  pulley   makes  110   R.  P.  M.    and  is  21   inches  in 
diameter  ;   what  should  be  the  size  of  the  driven  pulley  in  order  to 
make  385  R.  P.  M.  ?  Ans.  6  in. 

2.  The  main  shaft  of  a  certain  shop  makes  120   R.  P.  M.     It  is 
desired   to  have  the  countershaft  make   150  R.  P.  M.     There  are  on 
hand  pulleys  16  inches,  24  inches,  28  inches,  35  inches,  and  38  inches  in 
diameter.    Can  two  of  these  be  used,  or  must  a  new  pulley  be  ordered  ? 

Ans.  Use  the  28-inch  and  35-inch  pulleys. 

3.  The    pinion    (driver)    makes    174    R.  P.  M.    and    the    follower 
24  R.  P.  M. ;  how  many  teeth  must  the  pinion  have  if  the  follower  has 
87  teeth  ?  Ans.  12  teeth. 

4.  If  an  engine   flywheel  is    66    inches  in  diameter  and    makes 
160  R.  P.  M.,  what  must  be  the  diameter  of  the  pulley  on   the  main 
shaft  to  make  128  R.  P.  M.  ?  Ans.  82$  in. 

5.  What  is  the  pitch  diameter  of  a  gear  whose  circular  pitch  is 
1J  inches  and  has  28  teeth  ?  Ans.  11.14  in. 

6.  How  many  teeth  are  there  in  a  gear  whose  circular  pitch  is 
.7854  inch  and  which  is  23  inches  in  diameter  ?  Ans.  92  teeth. 

7.  What    is    the    circular    pitch    of    a    gear    whose    diameter    is 
20.372  inches  and  which  has  128  teeth  ?  Ans.  |  in. 

8.  What  is  the  pitch  diameter  of  a  gear-wheel  having  80  teeth, 
7  diametral  pitch  ?  Ans.  l^f  in. 

9.  What  is  the  over-all  diameter  of  a  gear-wheel  of  9  diametral 
pitch  and  47  teeth  ?  Ans.  5$  in. 


42  MACHINE  ELEMENTS.  §  5 

10.  What  is  the  diametral  pitch  of  a  gear-wheel  having  90  teeth 
and  an  outside  diameter  of  5f  inches  ?  Ans.  16  pitch. 

11.  How  many  teeth  of  20  diametral  pitch  can  be  cut  in  a  gear- 
wheel having  an  outside  diameter  of  2^  inches  ?  Ans.  40  teeth. 

12.  If  the  center-to-center  distance  of  two  gear-wheels  is  6  inches 
and  the  small  gear  is  to  make  4  revolutions  for  1  revolution  of  the 
large  gear,  what  must  be  the  diameter  of  the  large  gear  ? 

Ans.  9.6  in. 

13.  In  a  pair  of  gear-wheels,  the  driver  has  48  teeth  and  the  driven 
wheel  56  teeth.     If  the  driven  wheel  makes  98  revolutions  per  minute, 
how  many  revolutions  must  the  driver  make  ?  Ans.  114^  R.  P.  M. 

14.  In  a  train  of  gears,  the  drivers  have  16,  30,  24,  and  18  teeth, 
respectively;  the  followers  have  12,  24,  36,  40  teeth,  respectively.     If 
the  first  driver  makes  80  R.  P.  M.,  how  many  R.  P.  M.  will  the  last  fol- 
lower make  ?  Ans.  40  R.  P.  M. 


FIXED    AND    MOVABLE    PULLEYS. 

70.  Pulleys  are  also  used  for  hoisting  or  raising  loads,  in 
which  case  the  frame  that  supports  the  axle  of  the  pulley  is 
called  the  block. 

71.  A  flxed  pulley  is  one  whose  block  is  not  movable, 
as  in  Fig.  28.     In  this  case,   if  the  weight    W  be  lifted  by 


FIG.  29. 


pulling  down  P,  the  other  end  of  the  cord  W  will  evidently 
move  the  same  distance  upwards  that  P  moves  downwards; 
hence,  P  must  equal  W. 


§  5  MACHINE  ELEMENTS.  43 

72.  A  movable  pulley  is  one  whose  block  is  movable,  as 
in  Fig.  29.  One  end  of  the  cord  is  fastened  to  the  beam 
and  the  weight  is  suspended  from  the  pulley,  the  other  end 
of  the  cord  being  drawn  up  by  the  appli- 
cation of  a  force  P.  A  little  considera- 
tion will  show  that  if  Pacts  through  a 
certain  distance,  say  1  foot,  W  will  move 
through  half  that  distance,  or  6  inches ; 
hence,  a  pull  of  1  pound  at  P  will  lift 
2  pounds  at  W. 

The  same  would  also  be  true  if  the 
free  end  of  the  cord  were  passed  over  a 
fixed  pulley,  as  in  Fig.  30,  in  which  case 
the  fixed  pulley  merely  changes  the  direc- 
tion in  which  P  acts,  so  that  a  weight 
of  1  pound  hung  on  the  free  end  of  the 
cord  will  balance  2  pounds  hung  from 
the  movable  pulley. 

73.     A  combination  of  pulleys,  as 

shown  in  Fig.  31,  is  sometimes  used. 
In  this  case  there  are  three  movable 
and  three  fixed  pulleys,  and  the  amount 
of  movement  of  W,  owing  to  a  certain 
movement  of  P,  is  readily  found. 

It  will  be  noticed  that  there  are 
6  parts  of  the  rope,  not  counting  the 
free  end;  hence,  if  the  movable  block  be 
lifted  1  foot,  P  remaining  in  the  same 
position,  there  will  be  1  foot  of  slack  in 
each  of  the  6  parts  of  the  rope,  or 
6  feet  in  all.  Therefore,  Pmust  move 
6  feet  in  order  to  take  up  this  slack,  or 
P  moves  6  times  as  far  as  W.  Hence, 
FIG-  31-  1  pound  at  Pwill  support  6  pounds  at  W, 

since  the  force  multiplied  by  the  distance  through  which  it 
moves  equals  the  weigJit  multiplied  by  the  distance  through 
whicJi  it  moves.  It  will  also  be  noticed  that  there  are  three 
movable  pulleys,  and  that  3x2  =  6. 


44  MACHINE  ELEMENTS.  §  5 

LAW  OF  COMBINATION  OF  PULLEYS. 

74.  Rule  27. — In  any  combination  of  pulleys  wJiere  one 
continuous  rope  is  used,  a  load  on  the  free  end  will  balance  a 
weight  on  the  movable  block  as  many  times  as  great  as  itself 
as  there  are  parts  of  the  rope  supporting  the  load,  not  counting 
the  free  end,  assuming  that  there  are  no  friction  losses. 

The  above  law  is  good  whether  the  pulleys  are  side  by 
side,  as  in  the  ordinary  block  and  tackle,  or  whether  they 
are  in  line  and  beneath  one  another. 

EXAMPLE. — In  a  block  and  tackle  having  5  movable  pulleys,  how 
great  a  force  must  be  applied  to  the  free  end  of  the  rope  to  raise 
1,250  pounds,  assuming  that  there  are  no  friction  losses  ? 

SOLUTION.— Since  there  are  5  movable  pulleys,  there  must  be  10  parts 
of  the  rope  to  support  them.  Hence,  according  to  the  above  law,  a 
force  applied  to  the  free  end  will  support  a  load  10  times  as  great  as 

1  2^0 
itself,  or  the  force  =  =^  =  125  Ib.     Ans. 

75.  Frictional  Losses. — Owing  to  the  friction  of   the 
sheaves  and  the  friction  and  resistance  to  bending  of  the 
rope,  it  is  not  possible  to  move  a  weight  with  a  block  and 
tackle  by  applying  the  theoretical  force  calculated  by  the 
above    rule.      The    actual   pull   required    depends   on   such 
factors  as  the  condition  of  the  rope  and  the  friction  and 
number  of  the  sheaves.     By  experiment  it  has  been  deter- 
mined that  under  average  conditions  the  load  that  can  be 
raised  with  a  block  and  tackle  having  4  sheaves  will  average 
about  75  per  cent,   of  the  theoretical  load ;  with  5  sheaves, 
about  70  per  cent. ;  with  6  sheaves,  about  66  per  cent. ;  with 
7  sheaves,  about  63  per  cent. ;   and  with  8  sheaves,  about 
60  per  cent.     No  records  of  tests  with  a  larger  number  of 
sheaves  have  been  made  public,  but  it  is  fair  to  assume  that 
the  decrease  in  the  load  that  can  be  lifted  will  vary  in  about 
the  same  ratio. 

Letya  =  the  force  applied  at  the  free  end  of  the  rope; 

c  =  the  ratio  of  the  theoretical   to  the  actual  load 

corresponding  to  the  number  of  sheaves; 
m  =  the  number  of  movable  pulleys ; 
/  =  the  load  to  be  raised. 


§  5  MACHINE  ELEMENTS.  45 

Then,  the  force  that  is  actually  required  to  raise  a  given 
load,  may  be  found  approximately  by 

Rule  28.—  Divide  100  times  the  load  by  the  product  of 
twice  the  number  of  movable  pulleys  and  the  number  of  per 
cent,  corresponding  to  the  number  of  sheaves  used. 

f        100  / 

Or'  '-  =  8lS? 

EXAMPLE.  —  In  an  ordinary  block  and  tackle  having  three  movable 
pulleys,  as  shown  in  Fig.  31,  how  great  a  force  must  actually  be  applied 
to  raise  2,000  pounds  ? 

SOLUTION.  —  According  to  Art.  75,  c  may  be  taken  as  66,  there  being 
G  sheaves.  Then  applying  rule  28,  we  get 


76.  The  load  that  may  actually  be  raised  with  a  block 
and  tackle  will  be  approximately  given  by  the  following  rule: 

Rule  29.  —  Multiply  the  force  to  be  applied  to  the  free  end 
of  the  rope  by  t^v^ce  the  number  of  movable  pulleys  and  by 
the  number  of  per  cent,  corresponding  to  the  number  of  sheaves 
used.  Divide  the  product  by  100. 


EXAMPLE.  —  If  there  are  enough  men  pulling  on  the  free  end  of  the 
rope  of  a  block  and  tackle  having  two  movable  pulleys  to  exert  a  force 
of  250  pounds,  what  weight  may  they  expect  to  raise  ? 

SOLUTION.  —  There  being  4  sheaves,  by  Art.  75,  c  =  75.  Then  by 
rule  29, 


77.  If  the  free  end  of  the  hauling  rope  passes  first  around 
a  stationary  sheave,  as  in  Fig.  31,  it  does  not  make  any  dif- 
ference in  what  direction  in  the  plane  of  the  sheave  the  rope 
is  pulled.  If  it  passes  first  around  a  movable  sheave,  how- 
ever, as  in  Fig.  29,  the  pull  must  be  exerted  in  a  line  parallel 
to  the  line  of  action  of  the  resistance,  or  a  line  joining  the 


46 


MACHINE  ELEMENTS. 


centers  of  the  movable  and  stationary  sheaves,  in  order  to 
obtain  the  maximum  effect.  If  the  rope  pulls  on  the  mov- 
able sheave  at  an  angle,  its  effect  will  be  a  displacement 
sideways,  instead  of  straight  up. 

78.  The  Weston  differential  pulley  block,  more  com- 
monly known  as  a  chain  hoist,  is  shown  in  Fig.  32.  With 

the  help  of  this  device,  one 
man  can  hoist  a  load  far  be- 
yond what  can  be  done  with 
the  ordinary  block  and  tackle. 
There  are  two  pulleys  slightly 
different  in  size  alongside  of 
each  other  in  the  stationary 
block  and  rigidly  connected 
so  as  to  turn  together.  An 
endless  chain  passes  over 
both  pulleys  and  supports 
the  movable  block  in  one  of 
its  bights.  The  other  loop, 
or  bight,  is  free  and  forms 
the  hauling  part.  This  kind 
of  a  pulley  block  possesses 
the  valuable  property  that 
the  load  can  be  stopped  any- 
where by  simply  ceasing  to 
haul  on  the  hoisting  part  of 
the  free  loop. 

In  Fig.  33  the  differential  pulley  block  is  shown  in  dia- 
grammatic form.  Suppose  that  the  part  a  of  the  free  bight 
is  pulled  downwards,  that  is,  in  the  direction  of  the  arrow. 
Evidently,  the  upper  pulleys  will  both  turn  in  the  direction 
of  the  hands  of  a  watch,  as  indicated  by  the  curved  arrow. 
Then,  the  left-hand  part  of  the  bight  supporting  the  mov- 
able pulley  will  move  upwards,  and  the  right-hand  part 
downwards.  But  as  the  large  and  small  pulley  of  the  upper 
block  move  together,  it  follows  that  the  left-hand  side  of 
the  bight  supporting  the  movable  sheave  will  run  faster 


FIG.  32. 


FIG.  33. 


§  5  MACHINE  ELEMENTS.  47 

over  the  large  pulley  than  the  right-hand  side  of  the  bight 
will  run  downwards  over  the  small  pulley.  The  result  is 
that  the  upward  motion  of  the  movable  block  will  be  equal 
to  one-half  the  difference  of  the  distances  passed  over  in 
the  same  time  by  fixed  points  in  both  sides  of  the  bight. 

79.  Differential  chain  hoists  are  invariably  so  con- 
structed that  one  man  can  hoist  any  load  up  to  the  maximum 
capacity  of  the  hoists.  This  maximum  load  will  usually  be 
found  stamped  on  the  upper  block,  and  there  is,  hence,  no 
need  in  practice  of  calculating  the  force  that  must  be  exerted 
to  hoist  a  load  with  a  chain  hoist. 


EXAMPLES   FOB  PRACTICE. 

1.  In  a  block  and  tackle  with  two  movable  pulleys,  (a)  what  force 
will  be  required  to  lift  300  pounds,  assuming  that  there  are  no  friction 
losses  ?    (£)  What  force  may  be  expected  to  be  actually  required,  taking 
friction  into  account  ?  ( (a)    75  Ib. 

AnS'|(<*)     100  Ib. 

2.  A  man  weighing  120  pounds  throws  his  whole  weight  on  the  free 
end  of  the  rope  of  a  block  and  tackle  with  3  movable  pulleys.     What 
weight  can  he  raise,  (a)  assuming  that  there  are  no  friction  losses  ? 
(£)  taking  friction  into  account  ?  (  (a)    720  Ib. 

( (b)    475  Ib.,  about 


THE  INCLINED 

80.  An  inclined  plane  is  a  slope,  or  flat  surface,  making 
an    angle  with    another 

surface  or  base  plane. 

81.  Three  cases  may 
arise  in  practice  with  an 
inclined  plane  having  a 
horizontal     base     plane : 
(1)  Where  the  force  acts 
parallel  to  the  plane,  as 


48 


MACHINE  ELEMENTS. 


lib. 


in  Fig.  34.      (2)  Where  the  force  acts  parallel  to  the  base, 
as  in  Fig.  35.      (3)   Where  the  force  acts  at  an  angle  to  the 
,L  plane  or    to  the  base, 

as  in  Fig.  36. 


2lb.  (  o — } ^- —  82.  Relation  Be- 

tween Force  and 
Weight.— In  Fig.  34, 
the  relation  existing 
between  the  force  and 
the  weight  is  easily 
found.  The  weight  as- 
cends a  distance  equal 

to  cb,  or  the  height  of  the  inclined  plane,  while  the  force 

acts  through  a  distance 

equal    to    a  b,    or    the 

length    of   the    inclined 

plane.     Therefore, 

Rule  30.—  To  find 
t/te  force,  multiply  the 
iv eight  by  the  height  of 
the  plane  and  divide  the 
product  by  the  length 
of  the  plane.  To  find 
the  weight  that  can 
be  raised,  multiply  the 
force  by  the  length  of  the  plane  and  divide  this  product  by 
the  height  of  the  plane. 

Or,  let  P=  force; 

/  =  length  of  the  plane ; 
b  —  base  of  the  plane; 
//  =  height  of  plane; 
W  =  weight. 

Then,  P=^    and  W=       ' 


EXAMPLE  1.— The  length  of  an  inclined   plane  is  40  feet  and  its 
height  is  5  feet ;  what  force  is  required  to  sustain  a  weight  of  100  pounds  ? 


§  5  MACHINE  ELEMENTS.  49 

SOLUTION.— Applying  rule  3O,  we  have 

^  =  100_X_5  =  mlb      Ans 

EXAMPLE  2. — The  length  of  an  inclined  plane  is  8  feet  and  its  height 
is  15  inches;  what  weight  will  a  force  of  120  pounds  support  ? 

SOLUTION. — Applying  the  rule  just  given, 

^=iaqx96 

15 

83.  In  Fig.  35,  the  force  is  supposed  to  act  parallel  to 
the  base  for  any  position  of  W\  therefore,  while  W  is  mov- 
ing from  the  level  a  c  to  b,  or  through  the  height  c  b  of  the 
inclined  plane,  P  will  move  through  a  distance  equal  to 
the  length  of  the  base  a  c.  When  the  force  acts  parallel  to 
the  base,  we  have 

Rule  31. —  To  find  tJie  force  required,  multiply  the  weight  by 
theJicight  of  the  plane  and  divide  the  product  by  the  length  of 
the  base  of  the  plane.  To  find  the  weight  that  can  be  raised, 
multiply  the  force  by  the  length  of  the  base  of  the  plane  and 
divide  the  product  by  the  height  of  the  plane. 

Wh          ,„     Pb 

Or,  P=  — =—  and  JF=— =-. 

b  h 

EXAMPLE  1.— With  a  base  30  feet  long  and  a  height  of  6  feet,  what 
force  will  sustain  a  weight  of  75  pounds  ? 

SOLUTION. — By  rule  31, 

151b.     Ans. 

EXAMPLE  2. — A  force  of  12  pounds  sustains  a  weight  on  a  plane 
whose  base  is  6  feet  long  and  whose  height  is  18  inches;  find  the 
weight. 

SOLUTION. — Applying  the  rule  just  given,  we  have 
W  =  12  X  6  =  48  Ib.     Ans. 

For  Fig.  36,  no  rule  can  be  given.  The  ratio  of  the 
power  varies  as  the  position  of  W  varies ;  it  may  be  deter- 
mined by  the  principle  of  the  resolution  of  forces.  As  W 


50  MACHINE  ELEMENTS.  §  5 

shifts  its  position,  the  angle  that  its  cord  makes  with  the 
face  of  the  plane  varies,  and  the  magnitude  of  the  force  P 
depends  on  this  angle;  the  smaller  the  angle  the  less  is  P 
required  to  be  in  order  to  support  a  given  weight  on  a  given 
plane. 

84.  Applications  of  the  Inclined  Plane.— The  wedge 
is  a  movable  inclined  plane  that  in  various   modifications 

serves  for  many  different 
purposes.     In  the  form  of 
a  stake  wedge,  as  shown 
FIG.  37.  in   Fig.    37,    it  serves  to 

move  heavy  weights  or  to  separate  closely  joined  pieces  of 
material.  In  the  form  of  a  cold  chisel  or  knife,  it  serves  to 
sever  substances.  In  the  form  of  a  key,  it  serves  to  move 
the  brasses  in  the  different  pin-joint  connections  of  a  steam 
engine.  Familiar  examples  of  inclined  planes  about  machin- 
ery are  the  keys  used  for  fastening  a  crank  to  the  crank- 
shaft or  a  pulley  to  the  shaft;  in  some  designs  of  cross- 
heads,  the  cotters  securing  the  crosshead  to  the  piston  rod 
are  an  example  of  an  inclined  plane;  as  are  also  the  ad- 
justable shoes  in  many  designs  of  crossheads  for  steam 
engines. 

85.  Taper  of  Keys. — In  keys,  cotters,  and  stake  wedges, 
it    is  customary  to    express   the  inclinations   as    a   certain 
amount   per   foot,    or 

? — f 

taper  per  foot.    Thus,    H 

in   Fig.   38,    the  incli-  -*-| 1/  1 


nation  would  be  3  —  2      ^ IZ" 

=  1  inch  in  12  inches,  FIG.  as. 

and  the  taper  would  be,  as  there  are  12  inches  in  a  foot, 
1  inch  per  foot.  The  taper  per  foot  can  be  ascertained  by 
the  following  rule, 

where  a  =  the  large  diameter  in  inches; 

b  =  the  small  diameter  in  inches; 
/  =  the  length  in  inches; 
T  =  taper  in  inches  per  foot. 


MACHINE  ELEMENTS. 


51 


Rule  32. — Subtract  the  small  diameter  from  the  large 
diameter,  multiply  the  remainder  by  12,  and  divide  the  prod- 
uct by  the  length  of  the  key. 


Or, 


,r 

i  = 


EXAMPLE. — A  key  measures  4  inches  at  the  large  end  and  3  inches 
at  the  small  end ;  if  its  length  is  16  inches,  what  is  the  taper  per  foot  ? 
SOLUTION. — Applying  the  rule  just  given,  we  get 
r=(4-3)Xl2  =  f.nperft      Ang 

86.  Where  tapered  keys  are  used  for  adjusting  brasses, 
it  is  often  desirable  to  know  how  much  the  brasses  will  be 
closed  in  on  the  pin  by  a  given  movement  of  the  key,  or  how 
much  to  back  out  a  key  that  has  been  driven  home  until 
the  brasses  nip  the  pin,  the  clearance  between  the  brasses 
and  the  pin  necessary  for  satisfactory  running  having  been 
decided  on. 

Let       T  =  taper  in  inches  per  foot ; 

d  =  distance  the  key  is  driven  in  inches; 
x  —  amount  the  brass  is  moved  in  inches. 

Rule  33. —  To  find  the  amount  the  brass  has  been  moved, 
multiply  tlie  taper  by  the  distance  the  key  is  driven  and  divide 
the  product  by  12.  To  find  how  far  to  back  out  a  key, 
multiply  the  amount  the  brass  is  to  be  moved  back  by  12  and 
divide  by  the  taper. 


Or, 


Td 


EXAMPLE.— In  Fig.  39,   the  key  has  been  driven  home  until  the 
brasses  nip  the  pin;   it  has  been  deter-  T- 

mined  to  run  the  brasses  with  a  clearance 
of  ^  inch;  how  much  must  the  key  be 
backed  out,  its  taper  being  f  inch  per 
foot? 

SOLUTION. — By  rule  33, 


—  i  in.    Ans. 


FIG.  39. 


52  MACHINE  ELEMENTS.  §  5 

A  ready  way  of  measuring  the  distance  the  key  is  driven 
back  is  to  measure  from  the  top  surface  of  the  cotter  to  the 
upper  edge  of  the  key  while  the  key  is  home.  Add  the 
amount  the  key  is  to  be  driven  back,  and  move  it  until  a  rule 
shows  it  to  have  been  driven  the  right  amount.  This 
method  is  preferable  to  making  scriber  marks  on  the  flat 
sides  of  the  key,  as  there  is  no  liability  of  getting  mixed  up 
by  the  multiplicity  of  marks  sure  to  be  found  after  years 
of  service. 

87.  Keys  for  fastening  pulleys  to  shafts  in  order  to  cause 
them  to  rotate  together  belong  usually  to  one  of  the  three 
classes  shown  in  Fig.  40.  The  one  shown  at  a  is  a  concave 
key  that  is  hollowed  out  to  fit  the  shaft.  As  it  holds  the 
pulley  by  friction  alone,  it  is  only  suitable  for  light  work. 
In  case  such  a  key  slips,  its  holding  power  can  be  increased  a 
little  by  hollowing  out  its  concave  side  to  a  radius  smaller  than 


the  radius  of  the  shaft.  The  flat  key  is  shown  in  Fig.  40  at  b\ 
a  flat  surface  is  cut  on  the  shaft,  and,  consequently,  the  key 
is  more  effective  than  the  concave  key,  as  far  as  slipping 
under  a  rotative  strain  is  concerned.  The  sunk  key  shown  in 
Fig.  40  at  c  is  the  most  effective,  since  it  is  impossible  for 
the  pulley  to  turn  on  the  shaft  without  shearing  off  the  key. 
Keys  for  fastening  pulleys,  etc.  are  usually  given  a  taper  of 
from  i  to  ^  inch  per  foot. 


EXAMPLES  FOR  PRACTICE. 


1.  An  inclined  plane  is  30  feet  long  and  7  feet  high;  what  force  is 
required  to  roll  a  barrel  of  flour  weighing  196  pounds  up  the  plane,  the 
friction  being  neglected  ?  Ans.  45.7  Ib. 


MACHINE  ELEMENTS. 


53 


2.  What  is  the  taper  per  foot  of  a  stake  wedge  measuring  ^  inch  at 
the  small  end  and  ^  inch  at  the  large  end,  its  length  being  4  inches  ? 

Ans.  f£  in.  per  ft. 

3.  Given,  a  key  having  a  taper  of  1  inch  per  foot.     How  much  must 
it  be  backed  out  to  give  the  brasses  a  working  clearance  of  ^  inch  ? 

Ans.  fin. 


THE   SCREW. 

88.  A  screw  is  a  cylinder  having  a  helical  projection 
winding  around  its  circumference.     This  helix  is  called  the 

thread  of  the  screw. 
The  distance  that  a 
point  on  the  helix  is 
drawn  back    or   ad- 
vanced in  the  direc- 
tion of  the  length  of 
the  screw  during  one 
turn    is    called    the 
pitch  of  the  screw. 
The  screw  in  Fig.  41 
is  turned  in  a  nut  a 
by  means  of  a  force 
applied  at  the  end  of  the  handle  P.    For 
one  complete  revolution  of  the  handle, 
the  screw  will    be  advanced  lengthwise 
an  amount  equal  to  the  pitch.      If  the 
nut  is  fixed  and  a  weight  placed  upon 
the  end  of  the  screw,  as  shown,  it  will 
be  raised  vertically,  by  one  revolution 
of  the  screw,   a  distance  equal  to  the 
pitch.   During  this  revolution,  the  force 
at  P  will  move  through  a  distance  equal 
FIG.  4i.  to  the  circumference  of  a  circle  whose 

radius  is  P  F.     Therefore,  W  X  pitch  of  thread  =  P  X  cir- 
cumference of  circle  through  which  P  moves. 

89.  Owing  to  the  friction  between  the  nut  and  the  screw 
and  between  the  pivot  of  the  screw  and  its  cap,  the  weight 
that  can  be  raised  with  a  screw,   or  the  force  that  can  be 

a.  s.   I.— 17 


54  MACHINE  ELEMENTS.  §  5 

exerted  by  it  in  the  direction  of  its  axis,  is  far  less  than  that 
calculated  on  the  assumption  that  friction  does  not  exist. 
The  friction  depends  on  the  pitch,  the  condition  of  the 
thread  on  the  screw  and  in  the  nut,  and  the  nature  of  the 
lubricant. 

From  a  long  series  of  experiments,  Mr.  Wilfred  Lewis  has 
deduced  a  formula  for  determining  a  factor  by  which  to 
multiply  the  weight  that  could  be  raised  if  there  were  no 
friction,  so  as  to  determine  the  probable  actual  weight  that 
can  be  raised. 

Let   d  =  diameter  of  screw  in  inches; 
p  =  pitch  in  inches; 

E  =  the  number  by  which  to  multiply  the  theoretical 
weight  in  order  to  obtain  the  approximate  actual 
weight. 

Rule  34.  —  Add  the  diameter  of  the  screw  to  the  pitch  and 
divide  the  pitch  by  this  sum  ;  the  quotient  so  obtained  will  be 
tJie  factor  by  which  to  multiply  the  tJicoretical  weight  in  order 
to  determine  the  approximate  actual  weight.  • 

*-  :> 


EXAMPLE.—  With  a  screw  having  4  threads  to  the  inch  and  2  inches 
in  diameter,  what  is  the  factor  by  which  to  determine  how  much  of  the 
theoretical  load  can  probably  be  raised  ? 

SOLUTION.  —  Since  there  are  4  threads  per  inch,  the  pitch  is  1  -=-  4.  or 
i  inch.  Then  applying  rule  34,  we  get 


9O.  The  weight  that  can  be  lifted  when  friction  is 
neglected,  or  the  force  that  can  be  exerted  by  a  screw  when 
a  given  force  is  exerted  on  the  lever,  is  given  by  the  follow- 
ing rule, 

where  W=  weight  or  force  exerted  by  the  screw; 
P  =  force  applied  to  the  end  of  the  lever; 
/  =  pitch  ; 

r  =  distance  from  the  axis  of  the  screw  to  the  point 
of  application  of  the  force  P. 


§  5  MACHINE  ELEMENTS.  55 

Rule  35.  —  Multiply  tJie  force  applied  tot  lie  lever,  in  pounds, 
by  the  distance  of  its  point  of  application  from  the  axis  of  the 
screw,  in  inches,  and  multiply  this  product  by  6.2832  ;  divide 
the  result  by  the  pitch  in  inches  and  the  quotient  will  be  the 
theoretical  force  that  will  be  exerted  by  the  screw. 

Or,  '  w= 


91.  To  find  the  theoretical  force  that  must  be  applied  to 
the  lever  to  raise  a  given  weight  or  to  exert  a  given  force: 

Rule  36.  —  Multiply  the  weight  or  given  force  by  the  pitch 
and  divide  this  product  by  the  product  of  6.2832  and  the  dis- 
tance from  the  axis  of  the  screw  to  the  point  on  the  lever 
where  the  force  is  applied. 


6.2832  r 

92.  To  obtain  the  probable  actual  weight  that  may  be 
raised,  multiply  the  result  obtained  by  rule  35  by  the  value 
obtained  from  rule  34.  To  obtain  the  probable  force  that 
must  be  applied  to  the  lever  in  order  to  lift  a  given  weight, 
divide  the  result  obtained  from  rule  36  by  the  value  obtained 
from  rule  34. 

EXAMPLE  1.—  A  screw  jack  has  a  screw  If  inches  diameter  with 
5  threads  to  the  inch.  If  a  man  pulls  at  the  end  of  the  lever  14  inches 
from  the  axis  with  a  force  of  40  pounds,  what  weight  may  he  expect  to 
raise  ? 

SOLUTION.  —    5  threads  per  inch  —  £-inch  pitch. 

By  rule  35,  W  =  6-2832  *  40      If  _  17,593  pounds,  nearly. 

By  rule  34,  E=      ^      =  .1,  nearly. 

Then,  as  just  stated,  the  probable  weight  that  may  be  raised  is 
17,593  X  -1  =  1,759.3  Ib.     Ans. 

EXAMPLE  2.  —  The  pull  that  an  ordinary  man  can  exert  at  the  end  of 
a  lever  is  usually  taken  at  40  pounds  ;  would  it  be  possible  for  one  mail 


5fi 


MACHINE  ELEMENTS. 


to  raise  a  weight  of  3  tons  with  a  jack-screw  having  4  threads  to  the 
inch  and  2£  inches  in  diameter  when  the  longest  lever  available  is 
24  inches  ? 

SOLUTION. — According  to  rule  36,  the  theoretical  force  that  must 
be  exerted  is 


neaHy. 


By  rule  34, 


=  .091,  nearly. 


Then,  as  just  stated,  the  probable  actual  force  required  is  9.9. 
9.9  -r-  .091  =  109  pounds,  about,  or  more  than  a  single  man  may  be 
expected  to  exert  steadily.  Ans. 


SCREWS  USED  AS  FASTENINGS. 

93.  In  machine  construction,  screws  are  used  more  fre- 
quently as  fastening  devices  than  as  a  means  of  transmitting 
motion    or    producing  pressure.     Screws  used  as  fastening 
devices  may  be  divided  into  two  general  classes,  which  are 
distinguishable  from  each  other  by  the  form  of  thread  used : 
(1)    Metal  screius,   or  screws  intended  to  be  screwed   into 
metals.     (2)    Wood  screws,  or  screws  intended  for  wood. 

94.  Metal  screws  have  either  a  sharp  V  thread  or  the 
United  States  standard  thread  shown  in  Fig.  42.     In  the 


FIG.  42. 


latter,  as  also  in  the  sharp  V  thread,  the  sides  of  the  thread 
form  a  60°  angle.  In  the  United  States  standard  thread, 
the  thread  is  flattened  at  the  top  and  bottom,  as  shown  in 


§  5  MACHINE  ELEMENTS.  57 

the  figure.     The  advantage  of  this  is  that  it  makes  a  stronger 
screw  for  equal  pitches  and  diameter. 

95.  Metal  screws  are  divided  by  the  trade  into  machine 
bolts,  capscreivs,  and  machine  screws. 

96.  In  machine  bolts,  or  bolts,  for  short,  one  of  which 
is  shown  in  Fig.  43  (a),  the  shank  and  head  are  left  rough, 
or    just    as   they    come    from    the    bolt-heading    machine. 
Machine  bolts  are   regularly  made   in    lengths  varying  by 


half  inches  from  1£  to  8  inches.  Above  8  inches  and  up 
to  20  inches  the  length  varies  by  whole  inches.  The  length 
of  a  machine  bolt  is  always  measured  under  the  head,  as 
the  distance  /  in  Fig.  43  (a). 

97.  The  standard  diameters  in  which  machine  bolts  are 
regularly  made  are  ±,  TV,  f,  T\,  i,  TV,  f,  t,  £,  and    1   inch. 
Sizes  differing  from  these  are  special  and  can  only  be  obtained 
on  special  order.     By  the  diameter  of  a  bolt  is  always  meant 
its  diameter  over  the  top  of  the  thread.     Machine  bolts  can 
be  obtained  with  either  hexagonal  or  square  heads  and  with 
hexagonal  or  square  nuts. 

98.  Capscrews,  one  of   which  is  shown  in  Fig.  43  (£), 
are  bolts  with  either  hexagonal  or  square  heads;  they  have 
the  shanks  turned  and  the  heads  finished  all  over  by  milling 
and  turning.     Unlike  machine  bolts,  they  are  usually  fur- 
nished *without  nuts,  unless  especially  ordered. 

99.  Machine  bolts  and  capscrews  are  made  with  a  pitch 
of    thread   corresponding   to  the    United    States   standard 
thread,  as  shown  in  Table  II. 


58 


MACHINE  ELEMENTS. 
TABLE    II. 


STANDARD   THREADS   PER   INCH. 


Diameter. 
Inches. 

Threads  per 
Inch. 

Diameter. 
Inches. 

Threads  per 
Inch. 

i 

20 

H 

7 

Tt 

18 

if 

6 

1 

16 

H 

6 

A 

14 

if 

H 

i 

13 

if 

5 

iV 

12 

l* 

5 

1 

11 

2 

H 

1 

10 

2i 

4i 

-| 

9 

21- 

4 

i 

8 

** 

4 

H 

7 

3 

3£ 

The  only  deviation  from  this  table  of  threads  is  in  ^-inch 
square  and  hexagonal  capscrews,  where  the  usual  number 
of  threads  for  a  sharp  V  thread  is  12  per  inch  instead  of  13, 
as  called  for.  Capscrews  are  usually  furnished  with  a 
sharp  V  thread,  but  can  now  be  obtained  with  the  United 
States  standard  thread.  Machine  bolts  usually  have  the 
United  States  standard  thread  and  the  number  of  threads 
per  inch  called  for  in  the  table. 

1OO.  Machine  screws  are  small  screws  with  a  slotted 
head.  They  are  known  to  the  trade  as  flat  heads,  shown  in 

Fig.  44  (a) ;  round  heads,  shown 
in  Fig.  44  (b)\  button  heads, 
shown  in  Fig.  44  (c),  and  fillister 
heads,  shown  in  Fig.  44  (d). 

^  ^FIG  44"  ™  101'       SC1"eWS     f°r     UnHing 

metals  are  made  in  a  variety 
of  forms  that  differ  from  those  here  shown  to  suit  special 


§  5  MACHINE  ELEMENTS.  59 

conditions.  Some  of  these  are  setscrews,  carriage  bolts, 
tire  bolts,  boiler  patch  bolts,  elevator  bolts,  etc.  Most  of 
these  can  only  be  obtained  on  special  order,  although  some 
of  the  more  common  sizes  may  be  found  in  stock. 

1O2.  Wood  screws  are  either  flat  headed  or  round 
headed,  with  a  slotted  head.  When  wood  screws  have  a 
square  head  for  a  wrench,  instead  of  a  slot  for  a  screw- 
driver, they  are  known  as  lagscrews,  or  coach  screws. 


TELOCITY  RATIO  A1SD  EFFICIENCY. 

103.  By  the  term  velocity  ratio  is  meant  the  ratio  of 
the   distance   through  which  the  force   acts  to   the  corre- 
sponding distance  the  weight  moves.     Thus,  if  the  force  acts 
through  a  distance  of   12  inches  while  the  weight  moves 
1  inch,  the  velocity  ratio  is  12  to  1,  or  12,  that  is,  /'moves 
12  times  as  fast  as  W. 

If  the  velocity  ratio  be  known,  the  weight  that  any  machine 
will  raise,  under  the  assumption  that  there  are  no  frictional 
resistances,  can  be  found  by  multiplying  the  force  by  the 
velocity  ratio.  If  the  velocity  ratio  is  8.7  to  1,  or  8.7, 
W=  8.7  X  P,  since  Wxl  =  8.7. 

104.  In   any  machine,   the   force   actually  required  to 
raise  a  weight  or  overcome  a  resistance  is  always  greater 
than  the  quotient  obtained  by  dividing  this  weight,  or  resist- 
ance, by  the  velocity  ratio  of  the  machine. 

Thus,  if  there  were  no  friction,  a  machine  whose  velocity 
ratio  was  5  would,  by  an  application  of  100  pounds  of  force, 
raise  a  weight  of  500  pounds.  Now,  suppose  that  the  fric- 
tion in  the  machine  is  equivalent  to  10  pounds  of  force;  then, 
it  would  take  110  pounds  of  force  to  raise  500  pounds. 

If  in  the  above  illustration  friction  were  neglected,  we 
would  have  110  Ib.  x  5  =  550  Ib.  as  the  weight  that 
110  pounds  would  raise  ;  but  owing  to  the  frictional 


60  MACHINE  ELEMENTS.  §  5 

resistance,  it  only  raised  500  pounds.     Then,  we  have  for  the 

500 

ratio  between  the  two,  — -  =  .91.    That  is,  500  :  550 ::  .91  :  1. 
ooO 


1O5,  Efficiency.  —  The  ratio  between  the  weight  actu- 
ally raised  and  the  force  multiplied  by  the  velocity  ratio 
is  called  the  efficiency  of  the  machine.  For  example,  if 
the  weight  actually  raised  by  a  machine,  say  a  screw,  is 
1,600  pounds,  and  the  force  multiplied  by  the  velocity  ratio 
is  2,400  pounds,  the  efficiency  of  this  machine  is  |£##  =  .66f, 
or  66f  per  cent. 

EXAMPLE.  —  In  a  machine  having  a  combination  of  pulleys  and  gears, 
the  velocity  ratio  of  the  whole  is  9.75;  a  force  of  250  pounds  just  lifts 
a  weight  of  1,626  pounds;  what  is  the  efficiency  of  the  machine  ? 


SOLUTION.  —  Efficiency  =  H^—  TT-^H  =  -6671  or  66-71  Per  cent-    Ans- 


Since  the  total  amount  of  friction  varies  with  the  load, 
it  follows  that  the  efficiency  will  also  vary  for  different 
loads. 

106.  If  the  efficiency  of  a  machine  is  known,  the  force 
actually  required  to  raise  a  given  load  may  be  found  by 
dividing  the  load  by  the  product  of  the  velocity  ratio  of  the 
machine  and  the  efficiency.     Thus,   if   a    certain    machine 
has  a  velocity  ratio  of  10.6  and  its  efficiency  is  60  per  cent., 
the  force  that  must  actually  be  applied  to  raise  a  load  of 

O  A  A 

840  pounds  is  —  -          —  =  132.1  pounds,  nearly.      If  there 
-Lu.  o  X  •  oO 

have  been  no  losses  through  friction,  etc.,  the  force  required 
will  be  840  -j-  10.6  =  79.25  pounds,  nearly. 

107.  If  the  efficiency  is  known,  the  weight  that  a  cer- 
tain force  will  raise  may  be  found  by  multiplying  together 
the  force,  velocity  ratio,  and  the  efficiency.     Thus,  if  a  cer- 
tain machine  has  a  velocity  ratio  of  6^  and  an  efficiency 
of  78  per  cent.,  a  force  of  140  pounds  will  raise  a  weight  of 
140  X  6i  X  .78  =  709.9  pounds. 


§  5  MACHINE  ELEMENTS.  61 

EXAMPLES  FOR  PRACTICE. 

1.  The  distance  from  the  axis  of  a  screw  to  the  point  on  the  handle 
where  the  force  is  applied  is  12  inches.     The  screw  has  8  threads  per 
inch  and  is  1  inch  in  diameter,     (a)  What  force  is  necessary  to  raise  a 
weight  of  1,348  pounds,  assuming  that  there  is  no  friction  ?    (b)  What 
probable  force  will  be  required  ?  j  (a)    2.07  lb.,  nearly. 

iS'\(b}     19  lb.,  about. 

2.  What  weight  can  actually  be  raised  with  a  screw  jack  having  a 
screw  1±  inches  in  diameter  and  6  threads  per  inch,  when  a  man  is 
pulling  with  a  force  of  40  pounds  at  the  end  of  a  bar  16  inches  long  ? 

Ans.  2,838  lb.,  about. 

3.  In  example  2,  what  is  the  velocity  ratio  ?  Ans.  603,  nearly. 

4.  In  example  2,  what  is  the  efficiency  ?  Ans.  llf  per  cent. 


MECHANICS   OF    FLUIDS. 


HYDROSTATICS. 

1.  Hydrostatics  treats  of  liquids  at  rest  under  the  ac- 
tion of  forces. 

2.  Liquids  are  very  nearly  incompressible.     A  pressure 
of  15  pounds  per  square  inch  compresses  water  less  than 
inmnr  of  its  volume. 

3.  Hydrostatic  Pressure. — Fig.  1  represents  two  cylin- 
drical vessels  of  exactly  the  same  size.    The  vessel  a  is  fitted 
with  a  wooden  block  of  the  same 

size  as,  and  free  to  move  in,  the 
cylinder;  the  vessel  b  is  filled  with 
water,  whose  depth  is  the  same  as 
the  length  of  the  wooden  block 
in  a.  Both  vessels  are  fitted  with 
air-tight  pistons  P,  each  of  whose 
areas  are  10  square  inches. 

Suppose,  for  convenience,  that 
the  weights  of  the  pistons,  block, 
and  water  be  neglected,  and  that 
a  force  of  100  pounds  be  applied  to 
both  pistons.  The  pressure  per 
square  inch  will  be  ^y-  =  10  pounds. 
In  the  vessel  a  this  pressure  will 

be   transmitted  to  the  bottom  of  the   vessel,   and   will   be 
10  pounds  per  square  inch ;  it  is  easy  to  see  that  there  will 


For  notice  of  the  copyright,  see  page  immediately  following  the  title  page. 


MECHANICS  OF  FLUIDS. 


be  no  pressure  on  the  sides.  In  the  vessel  b  an  entirely 
different  result  is  obtained.  The  pressure  on  the  bottom 
will  be  the  same  as  in  the  other  case,  that  is,  10  pounds 
per  square  inch,  but,  owing  to  the  fact  that  the  molecules 
of  the  water  are  perfectly  free  to  move,  this  pressure  of 
10  pounds  per  square  inch  is  transmitted  in  every  direction 
with  the  same  intensity ;  that  is  to  say,  the  pressure  at  any 
point  c,  d,  e,  f,  g,  /i,  etc.,  due  to  the  force  of  100  pounds, 
is  exactly  the  same,  and  equals  10  pounds  per  square  inch. 

4.     The  foregoing  fact  may  be  easily  proved  experimen- 
tally by  means  of  an  apparatus  like  that  shown  in  Fig.  2. 

Let  the  area  of  the  piston 
a  be  20  square  inches;  of 
^,  7  square  inches ;  of  r, 
1  square  inch;  of  </, 
6  square  inches;  of  c, 
8  square  inches ;  and  of  /", 

4  square  inches. 

If  the  pressure  due  to 
the  weight  of  the  water  be 
neglected,  and  a  force  of 

5  pounds  be  applied  at   c 
(whose    area    is    1     square 
inch),     a      pressure      of 
5  pounds  per  square  inch 
will  be  transmitted  in    all 

FIG-  2-  directions,    and    in    order 

that  there  shall  be  no  movement,  a  force  of  6  X  5  =  30  pounds 
must  be  applied  at  d,  40  pounds  at  e,  20  pounds  at  f, 
100  pounds  at  a,  and  35  pounds  at  b. 

If   a  force  of    99  pounds  were  applied  to    a,  instead  of 

100  pounds,  the  piston  a  would  rise,  and  the  other  pistons 
#,  e,  (f,  e,  and  f  would  move  inwards ;  but,  if  the  force  applied 
to  a  were  100  pounds,  they  would  all  be  in  equilibrium.     If 

101  pounds  were  applied  at  a,  the  pressure  per  square  inch 
would  be  -Vjf  =  5.05  pounds,  which  would  be- transmitted  in 
all  directions;  and,  since  the  pressure  due  to  the  load  on  c  is 


§  6  MECHANICS  OF  FLUIDS.  3 

only  5  pounds  per  square  inch,  it  is  now  evident  that  the 
piston  a  will  move  downwards,  and  the  pistons  b,  c,  d,  e, 
and  f  will  be  forced  outwards. 

5.  Pascal's  Law. — The  whole  of  the  preceding  may  be 
summed  up  as  follows: 

The  pressure  per  unit  of  area  exerted  anywhere  on  a  liquid 
is  transmitted  undiminished  in  all  directions  and  acts  with 
the  same  force  on  all  surfaces,  in  a  direction  at  right  angles 
to  those  surfaces. 

This  law,  first  discovered  by  Pascal,  and  accordingly 
named  after  him,  is  the  most  important  one  in  hydrostatics. 
Its  meaning  should  be  thoroughly  understood. 

EXAMPLE. — If  the  area  of  the  piston  e,  in  Fig.  2,  were  8.25  square 
inches  and  a  force  of  150  pounds  were  applied  to  it,  what  forces  would 
have  to  be  applied  to  the  other  pistons  to  keep  the  water  in  equilibrium, 
assuming  that  their  areas  were  the  same  as  given  before  ? 

SOLUTION. —     Q~CJK  =  18.182  pounds  per  square  inch,  nearly. 

20  X  18.182  =  363.64    Ib.  =  force  to  balance  a. 
7  X  18.182  =  127.274  Ib.  =  force  to  balance  b. 
1  X  18.182  =    18.182  Ib.  =  force  to  balance  c.    \  Ans. 
6  X  18.182  =  109.092  Ib.  =  force  to  balance  d. 
4  X  18.182  =    72.728  Ib.  =  force  to  balance  /. 

6.  The  pressure  due  to  the  weight  of  a  liquid  may  be 

•tftivtiiwr/lv     i/S>Tfi/rr/7<:    nr  siffftctisp 


. 

downwards,  upwards,  or  sideivise. 


7.  Downward  Pressure.  —  In  Fig.  3  the  pressure  on 
the  bottom  of  the  vessel  a  is,  of  course,  equal  to  the  weight 
of  the  water  it  contains.  If  the  area  of  the  bottom  of  the 
vessel  b  and  the  depth  of  the  liquid  contained  in  it  are  the 
same  as  in  the  vessel  a,  the  pressure  on  the  bottom  of  b  will 
be  the  same  as  on  the  bottom  of  a.  Suppose  the  bottoms 
of  the  vessels  a  and  b  are  6  inches  square,  and  that  the 
part  c  d  in  the  vessel  b  is  2  inches  square,  and  that  they 
are  filled  with  water.  Then,  since  1  cubic  foot  of  water 
weighs  62.5  pounds,  the  weight  of  1  cubic  inch  of  water  is 

'  pound  —  .03617  pound.  The  number  of  cubic  inches 
l,72b 


MECHANICS  OF  FLUIDS. 


inrt  =  6x6x24  =  864  cubic  inches;  and  the  weight  of  the 
water  is  864  X  .03617  =  31.25  pounds.  Hence,  the  total 
pressure  on  the  bottom  of  the  vessel  a  is  31.25  pounds, 

31  25 
or    g-^-g  =  -868    pound    per 

square  inch. 

The  pressure  in  b  due  to 
the  weight  contained  in  the 
part  £<ris6x6xlOX  .03617 
=  13.02  pounds. 

The  weight  of  the  part  con- 
tained in  c  d  is  2  X  2  X  14 
X  .03617  =  2,0255  pounds, 
and  the  weight  per  square 

,   .     2.0255 
inch  of  area  in  c  d  is  — - — 

=  .5064  pound. 


8.  According  to  Pascal's  law,  this  weight  (pressure) 
is  transmitted  equally  in  all  directions,  therefore,  an  extra 
weight  of  .5064  pound  is  imposed  on  every  square  inch  of  the 
bottom  of  b  c\  the  area  of  this  is  6  X  6  =  36  square  inches, 
and  the  pressure  on  it  due  to  the  water  contained  in  c  d 
is,  therefore,  36  X  .5064  =  18.23  pounds;  thus,  we  have  a 
total  pressure  on  the  bottom  of  vessel  hot  13.02  +  18.23 
=  31.25  pounds,  the  same  as  in  vessel  a.  As  a  result  of  the 
above  law,  there  is  also  an  upward  pressure  of  .5064  pound 
acting  on  every  square  inch  of  the  top  of  the  enlarged 
part  b  c. 

If  an  additional  pressure  of  10  pounds  per  square  inch 
were  applied  to  the  upper  surface  of  both  vessels,  the  total 
pressure  on  each  bottom  would  be  31.25  +  (6  X  6  X  10) 
=  31.25  +  360  =  391.25  pounds. 

If  this  pressure  were  to  be  obtained  by  means  of  a  weight 
placed  on  each  piston  (as  shown  in  Fig.  1  at  a  and  b},  we 
would  have  to  put  a  weight  of  6  X  6  X  10  =  360  pounds  on 
the  piston  in  vessel  a,  Fig.  3,  and  one  of  2  X  2  X  10  =  40  pounds 
on  the  piston  in  vessel  b. 


MECHANICS  OF  FLUIDS. 


9.  The  general  law  for  tlie  downward  pressure  on 
the  bottom  of  any  vessel: 

Rule  1. —  The  pressure  on  the  bottom  of  a  vessel  containing 
a  fluid  is  independent  of  the  shape  of  the  vessel,  and  is  equal 
to  the  weigJit  of  a  column  of  tJie  fluid,  the  area  of  whose  base 
is  equal  to  that  of  the  bottom  of  the  vessel  and  whose  altitude 
is  the  distance  between  the  bottom  and  the  upper  surface  of 
the  fluid,  increased  by  the  pressure  per  unit  of  area  on  the 
upper  surface  of  the  fluid  multiplied  by  the  area  of  the  bottom 
of  the  vessel,  in  case  there  is  any  pressure  on  the  surface. 

10.  vSuppose  that  the  vessel  b,  in  Fig.  3,   were  inverted, 
as  shown  in  Fig.  4,  the  pressure  on  the  bottom 

would  still  be  .868  pound  per  square  inch,  but 
it  would  require  a  weight  of  3,490  pounds  on  a 
piston  at  the  upper  surface  to  make  the 
pressure  on  the  bottom  391.25  pounds, 
instead  of  a  weight  of  40  pounds,  as  in  the 
other  case. 

EXAMPLE.— A  vessel  filled  with  salt  water,  weigh- 
ing .037254  pound  per  cubic  inch,  has  a' circular 
bottom  13  inches  in  diameter.  The  top  of  the 
vessel  is  fitted  with  a  piston  3  inches  in  diameter,  on 
which  is  laid  a  weight  of  75  pounds;  what  is  the 
total  pressure  on  the  bottom,  if  the  depth  of  the  water  is  18  inches  ? 

SOLUTION. — Applying  the  rule,  we  have 

13  X  13  X  -7854  X  18  X  .037254  =  89.01  pounds,  the  pressure  due  to 
the  weight  of  the  water. 

5 o    ,   VOKA  =  10-61  pounds  per  square  inch,  due  to  the  weight  on 

O  X  o  X  •  <oO4 

the  piston. 

13  X  13  X  .7854  X  10.61  =  1,408.29  pounds. 

1,408.29  +  89.01  =  1,497.3  Ib.  =  total  pressure.     Ans. 

11.  Upward    Pressure. — In    Fig.    5    is    represented   a 
vessel  of   exactly  the   same  size  as  that  shown  in   Fig.  4. 
There  is  no  upward  pressure  on  the  surface  c  due  to  the 
weight  of  the  water  in  the  large  part  c  d,  but  there  is  *an 
upward  pressure  on  c  due  to  the  weight  of  the  water  in  the 


MECHANICS  OF  FLUIDS. 


*l 


FIG.  5. 


small  part  b  c.  The  pressure  per  square  inch  due  to  the 
weight  of  the  water  in  b  c  was  found  to  be  .5064  pound; 
the  area  of  the  upper  surface  c  of  the  large  part  c  d  is 
(6  X  6)  -  (2  X  2)  =  36  —  4  —  32  square  inches,  and  the 
total  upward  pressure  due  to  the  weight  of 

r—  ^=-  the  water  is  .5064  X  32  =  16.2  pounds. 
If  an  additional  pressure  of  10  pounds  per 
square  inch  were  applied  to  a  piston  fitting 
in  the  top  of  the  vessel,  the  total  upward 
pressure  on  the  surface  c  would  be  16.2 
+  (32  X  10)  =  336.2  pounds. 

1 2.     General  law  for  upward  pressure  : 

Rule  2. —  The  upward  pressure  on  any  sub- 
merged horizontal  surface  equals  the  weight 
of  a  column  of  the  liquid  whose  base  has  an 
area  equal  to  the  area  of  the  submerged  sur- 
face and  whose  altitude  is  the  distance 
between  the  submerged  surface  and  the  upper  surface  of  the 
liquid,  increased  by  the  pressure  per  unit  of  area  on  the  upper 
surface  of  the  fluid  multiplied  by  the  area  of  the  submerged 
surface,  in  case  of  any  pressure  on  the  upper  surface. 

EXAMPLE. — A  horizontal  surface  6  inches  by  4  inches  is  submerged 
in  a  vessel  of  water  26  inches  below  the  upper  surface ;  if  the  pressure 
on  the  water  is  16  pounds  per  square  inch,  what  is  the  total  upward 
pressure  on  the  horizontal  surface  ? 

SOLUTION.— Applying  the  rule,  we  get  6  X  4  X  26  X  .03617  =  22.57 
pounds  for  the  upward  pressure  due  to  the  weight  of  the  water,  and 
6  X  4  X  16  =  384  pounds  for  the  upward  pressure  due  to  the  outside 
pressure  of  16  pounds  per  square  inch. 

Therefore,  384  +  22.57  =  406.57  Ib.  =  the  total  upward  pressure. 

Ans. 

13.  Lateral  (Sidewise)  Pressure. — Suppose  the  top  of 
the  vessel  shown  in  Fig.  6  is  10  inches  square  and  that  the 
projections  at  a  and  b  are  1  inch  square. 

The  pressure  per  square  inch  on  the  bottom  of  the  vessel 
due  to  the  weight  of  the  liquid  is  1  X  1  X  18  X  the  weight 
of  a  cubic  inch  of  the  liquid. 


MECHANICS  OF  FLUIDS. 


The  pressure  at  a  depth  equal  to  the  distance  of  the  upper 
surface  of  b  is  1  X  1  X  17  X  the  weight  of  a  cubic  inch  of  the 
liquid. 

Since  both  of  these 
pressures  are  transmit- 
ted in  every  direction, 
they  are  also  transmit- 
ted laterally  (sidewise), 
and  the  pressure  per 
unit  of  area  on  the 
projection  b  is  a  mean 
between  the  two,  and 

equals  1  X  1  X  ^±^ 
X  the  weight  of  a  cubic 

..        ,  r- 10.   o. 

inch  of  the  liquid. 

To  find  the  lateral  pressure  on  the  projection  a,  imagine 
that  the  dotted  line  c  is  the  bottom  of  the  vessel ;  then,  the 
conditions  would  be  the  same  as  in  the  preceding  case, 
except  that  the  depth  is  not  so  great. 

The  lateral  pressure  per  square  inch  on  a  is  thus  seen  to 


be  1  X  1  X 


11       12 


X  the  weight  of  a  cubic  inch  of  the  liquid. 


14.     General  law  for  lateral  pressure : 

Rule  3. —  The  pressure  on  any  vertical  surface  due  to  the 
weight  of  the  liquid  is  equal  to  the  weight  of  a  column  of  the 
liquid  whose  base  has  the  same  area  as  the  vertical  surface 
and  whose  altitude  is  the  depth  of  the  center  of  gravity  of  the 
'vertical  surface  below  the  upper  surface  of  the  liquid. 

Any  additional presstire  is  to  be  added,  as  in  the  previous  cases. 

EXAMPLE.— A  well  3  feet  in  diameter  and  20  feet  deep  is  filled  with 
water ;  what  is  the  pressure  on  a  strip  of  wall  1  inch  wide,  the  top  of 
which  is  1  foot  from  the  bottom  of  the  well  ?  What  is  the  pressure  on 
the  bottom  ?  What  is  the  upward  pressure  per  square  inch  2  feet 
6  inches  from  the  bottom  ? 

SOLUTION. — Applying  the  rule,  the  area  of  the  strip  is  equal  to 
its  length  (=  circumference  of  well)  multiplied  by  its  height.  .The 
length  =  36  x  3.1416  =  113.1  inches;  height  =  1  inch;  hence,  area  of 
ff.  $.  I.-18 


8 


MECHANICS  OF  FLUIDS. 


strip  =  113.1  X  1  =  H3.1  square  inches.  Depth  of  center  of  gravity 
of  strip  =  (20  —  1)  feet  +  |  inch,  the  half  width  of  strip  =  228£  inches. 
Consequently,  113.1  X  228.5  X  .03617  =  934.75  Ib.  =  the  total  pressure 
on  the  strip.  Ans. 

93475  =  8.265  Ib., 


The  pressure  on  each  square  inch  of  the  strip  is 
nearly.     Ans. 

36  X  36  X  .7854  X  20  X  12  X  .03617  =  8,836  Ib.,   the  pressure   on   the 
bottom.     Ans. 

20  -  2.5  =  17.5.      1  X  17.5  X  12  X  .03617  =  7.596  Ib.,  the  upward  pres- 
sure per  square  inch  2  ft.  6  in.  from  the  bottom:     Ans. 


15. 

Fig.  7. 


The   effects   of   lateral   pressure   are   illustrated   in 
In  the  figure,  e  is  a  tall  vessel  having  a  stop-cock 

near  its  base  and  ar- 
ranged to  float  upon 
the  water,  as  shown. 
When  this  vessel  is 
filled  with  water,  the 
lateral  pressures  at 
any  two  points  of  the 
surface  of  the  vessel 
opposite  each  other 
are  equal.  Being 
equal  and  acting  in 
opposite  directions, 
they  destroy  each 
other,  and  no  motion 
FIG>  7<  can  result ;  but,  if  the 

stop-cock  is  opened,  there  will  be  no  resistance  to  the  pressure 
acting  on  that  part,  and  the  water  will  flow  out ;  at  the 
same  time,  the  reaction  of  the  jet  issuing  from  the  stop-cock 
causes  the  vessel  to  move  backwards  through  the  water  in  a 
direction  opposite  to  that  of  the  issuing  jet. 

16.  Liquids  Influenced  by  Gravity. — Since  the  pres- 
sure on  the  bottom  of  a  vessel  due  to  the  weight  of  the 
liquid  is  dependent  only  on  the  height  of  the  liquid  and 
not  on  the  shape  of  the  vessel,  it  follows  that  if  a  vessel  has 
a  number  of  radiating  tubes  (see  Fig.  8)  the  water  in  each 


§  6  MECHANICS  OF  FLUIDS.  9 

tube  will  be  on  the  same  level,  no  matter  what  may  be  the 
shape  of  the  tubes.  For,  if  the  water  were  higher  in  one 
tube  than  in  the  others,  the  downward  pressure  on  the 
bottom  due  to  the  height  of  the  water  in  this  tube  would  be 
greater  than  that  due  to  the  height  of  the  water  in  the  other 
tubes.  Consequently,  the  upward  pressure  would  also  be 
greater;  the  equilibrium  would  be  destroyed  and  the  water 


would  flow  from  this  tube  into  the  vessel  and  rise  in  the  other 
tubes  until  it  was  at  the  same  level  in  all,  when  it  would  be 
in  equilibrium.  This  principle  is  expressed  in  the  familiar 
saying,  water  seeks  its  level. 

An  application  of  this  principle  is  the  glass  water  gauge 
used  for  showing  the  level  of  the  water  in  a  steam  boiler 
or  tank. 

17.  The  above  principle  explains  why  city  water  reser- 
voirs are  located  on  high  elevations  and  why  water  on  leaving 
the  hose  nozzle  spouts  so  high. 

If  there  were  no  resistance  by  friction  and  air,  the  water 
would  spout  to  a  height  equal  to  the  level  of  the  water  in  the 
reservoirs.  If  a  long,  vertical  pipe,  whose  length  was  equal 
to  the  vertical  distance  between  the  nozzle  and  the  level  of 
the  water  in  the  reservoir,  were  attached  to  the  nozzle,  the 


10  MECHANICS  OF  FLUIDS.  §  6 

water  would   just  reach  the  end  of   the  pipe.      If    the  pipe 
were  lowered  slightly,  the  water  would  trickle  out. 

Fountains,  canal  locks,  and  artesian  wells  are  examples  of 
the  application  of  this  principle. 

EXAMPLE.— The  water  level  in  a  city  reservoir  is  150  feet  above  the 
level  of  the  street ;  what  is  the  pressure  of  the  water  per  square  inch 
on  the  hydrant  ? 

SOLUTION.— Applying  rule  1,  1  X  150  X  12  X  .03617  =  65.106  Ib.  per 
sq.  in.  Ans. 

NOTE. — In  measuring  the  height  of  the  water  to  find  the  pressure 
that  it  produces,  the  vertical  height,  or  distance,  between  the  level 
of  the  water  and  the  point  considered  is  always  taken.  This  vertical 
height  is  called  the  head. 

The  weight  of  a  column  of  water  1  inch  square  and  1  foot  high 
is  62.5  -f- 144  =  .434  pound,  nearly.  Hence,  if  the  depth  (head)  be  given, 
the  pressure  per  square  inch  may  be  found  by  multiplying  the  depth  in 
feet  by  .434.  The  constant  .434  is  the  one  ordinarily  used  in  practical 
calculations. 

18.  In  Fig.  9,  let  the  area  of  the  piston  a  be  1  square 
inch;  of  b,  40  square  inches.  According  to  Pascal's  law, 
1  pound  placed  on  a  will  balance  40  pounds  placed  on  b. 

Suppose  that  a  moves  downwards  10  inches,  then  10  cubic 
inches  of  water  will  be  forced  into  the  tube  b.  This  will  be 
distributed  over  the  entire  area  of  the 
tube  £,  in  the  form  of  a  cylinder  whose 
cubical  contents  must  be  10  cubic  inches, 
whose  base  has  an  area  of  40  square 
inches,  and  whose  altitude  must  be  ££ 
=  \  inch;  that  is,  a  movement  of 
10  inches  of  the  piston  a  will  cause  a 
movement  of  £  inch  in  the  piston  b. 
This  is  another  illustration  of  the  well- 
known  principle  of  machines:  The 
force  multiplied  by  the  distance  tJirough 
FIG-  9-  whicli  it  acts  equals  the  weight  multi- 

plied by   the    distance   through   which    it    moves,    since,    if 
1  pound  on  the  piston  a  represents  the  force  /*,  the  equiva- 
lent weight    W  on  b  may  be  obtained   from  the    equation 
IV  X  i  =  /3X  10,  whence  W=  40/}  =  40  pounds. 
Another  familiar  fact  is  also  recognized,  for  the  velocity 


§  0  MECHANICS  OF  FLUIDS.  11 

ratio  of  P  to  W  is  10  :  \,  or  40;  and  since  in  any  machine 
the  weight  equals  the  force  multiplied  by  the  velocity  ratio, 
W—Px^,  and  when  P=  1,  W  =•  40.  An  interesting 
application  of  this  principle  is  the  hydraulic  jack,  by  the  aid 
of  which  one  man  can  lift  a  very  large  load. 

19.  A  Watson -Stillman  hydraulic  jack  is  shown  in 
section  in  Fig.  10  (a).  In  this  illustration,  a  is  a  lever  that 
is  depressed  by  the  operator  when  he  desires  to  raise  the 
load.  This  lever  freely  fits  a  rectangular  hole  or  socket  in 
a  shaft  /;  that  extends  clear  through  the  ram  head.  The 
shaft  b  carries  the  crank  c  to  which  the  piston  d  is  hinged. 
A  small  valve  is  placed  in  the  center  of  the  piston  and 
serves  to  open  or  close  communication  between  the  water 
space  above  and  below  the  piston.  The  lower  end  of  the 
ram  is  fitted  with  a  small  check  valve,  which  normally  is 
held  closed  by  the  small  spring  shown  beneath  the  valve. 
The  inside  of  the  jack  is  filled  with  a  mixture  of  alcohol  and 
water,  about  2  parts  of  alcohol  to  3  parts  of  water,  which 
prevents  the  freezing  of  the  liquid  when  the  jack  is  used  in 
cold  weather.  The  operation  is  as  follows : 

The  jack  being  clear  down,  as  shown  in  Fig.  10  (a),  it  is 
placed  under  the  load  and  the  operator  alternately  depresses 
and  raises  the  lever  through  its  full  range,  the  motion  being 
limited  by  a  projection  on  the  under  side  of  the  lever.  The 
slightest  depression  of  the  lever  closes  the  valve  in  the  pis- 
ton. The  downward  movement  of  the  piston  continuing, 
the  water  between  the  bottom  of  the  piston  and  the  end  of 
the  ram  is  subjected  to  a  pressure  that  opens  the  valve  in  the 
end  of  the  ram.  This  allows  the  water  under  pressure  to 
pass  into  the  space  beneath  the  ram.  The  pressure  so  trans- 
mitted to  the  lower  end  of  the  ram  forces  it  upwards.  The 
lever  having  been  depressed  to  its  limit  is  again  raised. 
This  raises  the  piston,  causes  the  piston  valve  to  open,  and 
allows  the  water  above  the  piston  to  flow  beneath  it.  The 
pressure  of  the  water  below  the  ram  promptly  closes  the  ram 
valve  as  soon  as  the  piston  commences  to  move  upwards. 
The  operation  may  now  be  repeated  as  often  as  required. 


12 


MECHANICS  OF  FLUIDS. 


As  will  readily  be  seen,  the  ram  cannot  descend  under  the 
weight  of  the  load,  as  the  only  water  passage  leading  from 


FIG.  10. 


the  space  below  the  ram  is  closed  by  the  ram  valve,  which 
is  kept  tightly  closed  by  the  pressure  of  the  water  beneath 


§  6  MECHANICS  OF  FLUIDS.  13 

it  and  also  by  the  spring.  Hence,  in  order  to  lower  the 
ram,  this  ram  valve  must  be  opened  by  some  means.  As 
previously  mentioned,  the  lever  has  a  projection  on  its  under 
side  to  limit  its  travel  and  the  travel  of  the  piston.  This 
projection  is  made  of  such  a  length  that  when  the  lever  is 
down  as  far  as  it  will  go,  the  piston  is  still  a  short  distance 
from  the  upper  end  of  the  ram  valve  stem. 

In  order  to  lower  the  ram,  the  lever  is  withdrawn  from  its 
socket  and  inserted  again  with  the  projection  upwards.  It 
can  now  be  depressed  far  enough  to  cause  the  end  of  the 
piston  to  strike  and  open  the  ram-head  valve,  as  shown  in 
Fig.  10  (d),  where  the  parts  of  the  jack  are  shown  in  the 
relative  positions  occupied  while  lowering.  But  before  the 
water  in  the  bottom  of  the  ram  can  escape  to  the  top, 
the  valve  in  the  piston  must  also  be  opened.  This  is  accom- 
plished by  a  heavy  spring  that  forces  the  sleeve  e  downwards. 
The  lower  end  of  the  sleeve  has  a  cotter  working  in  a  slot  in 
the  piston  rod;  the  cotter  bearing  against  the  top  of  the 
valve  forces  the  latter  downwards.  The  sleeve  is  connected 
to  the  crank  c  by  the  so-called  lowering  wire  f,  in  such  a 
manner  that  the  cotter  will  not  strike  the  piston  valve  while 
the  jack  is  raising  the  load.  The  ram  will  descend  as  long 
as  the  lever  is  kept  hard  down ;  it  can  be  stopped  instantly 
by  raising  the  lever  slightly. 

2O.  If  the  area  of  the  piston  is  1  square  inch  and  the 
area  of  the  ram  is  10  square  inches,  the  velocity  ratio  will  be 
10.  If  the  length  of  the  lever  between  the  hand  of  the  oper- 
ator and  the  fulcrum  is  10  times  the  length  between  the 
fulcrum  and  the  piston,  the  velocity  ratio  of  the  lever  will  be 
10,  and  the  total  velocity  ratio  of  the  hand  to  the  piston 
will  be  10  X  10  =  100.  Taking  the  weight  of  the  operator 
at  150  pounds,  his  whole  weight  to  be  thrown  on  the  lever, 
the  weight  that  can  be  raised  is  150  X  100  =  15,000  pounds, 
or  7£  tons.  But  if  the  average  movement  of  the  hand  is 
4  inches  per  stroke,  it  will  require  J-^  =  25  strokes  to  raise 
the  load  on  the  jack  a  distance  of  1  inch,  and  it  is  again 
seen  that  what  is  gained  in  force  is  lost  in  speed. 


14  MECHANICS  OF  FLUIDS.  §  6 

21.  Other  applications  of  this  principle  are  seen  in  vari- 
ous hydraulic  machines  used   in  boiler  shops.      A  familiar 
example  is  the  hand  test  pump  used  in  testing  boilers  under 
hydraulic  pressure. 

EXAMPLE  1. — A  vertical  cylinder  is  tested  for  the  tightness  of  its 
heads  by  filling  it  with  water.  A  pipe  whose  inside  diameter  is  ^  inch 
and  whose  length  is  20  feet  is  screwed  into  a  hole  in  the  upper  head 
and  is  then  filled  with  water.  What  is  the  pressure  per  square  inch  on 
each  head  if  the  cylinder  is  40  inches  in  diameter  and  60  inches  long  ? 

SOLUTION.—    402  X  .7854  =  1,256.64  square  inches  =  area  of  head. 
1  X  60  X  .03617  =  2.17  pounds  pressure  per  square  inch  on  the  bottom 
head  due  to  the  weight  of  the  water  in  the  cylinder. 

(i)2  X  -7854  =  .04909  square  inch,  the  area  of  the  pipe. 

.04909  X  20  X  12  X  .03617  =  .426  pound  =  the  weight  of  water  in  the 
pipe  =  the  pressure  on  a  surface  area  of  .04909  square  inch. 

The  pressure  per  square  inch  due  to  the  water  in  the  pipe  is 
X  -426  =  8.68  Ib.  per  sq.  in.  upon  the  upper  head.     Ans. 

The  total  pressure  per  square  inch  on  the  lower  head  is  8.68  +  2.17 
=  10.85  Ib.  Ans. 

EXAMPLE  2.— In  the  other  example,  if  the  pipe  be  fitted  with  a 
piston  weighing  £  pound  and  a  5-pound  weight  be  laid  on  it  what  is 
the  pressure  per  square  inch  on  the  upper  head  ? 

SOLUTION. — In  addition  to  the  pressure  of  .426  pound  on  the  area  of 
.04909  square  inch,  there  is  now  an  additional  pressure  on  this  area  of 
5  -h  i  =  5.25  pounds,  and  the  total  pressure  on  this  area  is  .426  +  5.25 

=  5.676  pounds.  X  5.676  =  115.6  Ib.  =  the  pressure  per  square 

.  04«7oy 
inch.     Ans. 

22.  When  calculating  the  weight  that  can  be  raised  with 
a  hydraulic  jack,  no  allowance  was  made  for  the  power  lost 
in  overcoming  the  friction  between  the  cup  leathers  of  the 
piston  and  the  ram,  and  the  cup   leather  of   the  ram  and 
the  cylinder;  this  varies  according  to  the  condition  of  the 
leathers,  and,   of  course,   the  smoothness  of   the   ram   and 
cylinder;  when  the  leathers  are  in  good  condition,  the  loss  is 
about  5  per  cent,  of  the  total  pressure  on  the  ram ;  when  the 
leathers  are   old,  stiff,   and  dirty,  the  loss  may  amount  to 

15  per  cent,  or  more. 


MECHANICS  OF  FLUIDS. 


15 


SPECIFIC  GRAVITY. 

23.  The  specific  gravity  of  a  body  is  the  ratio  between 
its  weight  and  the  weight  of  a  like  volume  of  water. 

24.  Since  gases  are  so  much  lighter  than  water,  it  is 
usual  to  take  the   specific  gravity  of   a   gas   as   the   ratio 
between  the  weight  of  a  certain  volume  of  the  gas  and  the 
weight  of  the  same  volume  of  air. 

EXAMPLE.— A  cubic  foot  of  cast  iron  weighs  450  pounds;  what  is  its 
specific  gravity,  a  cubic  foot  of  water  weighing  62.42  pounds  ? 
SOLUTION. — According  to  the  definition, 
450 


62.42 


=  7.21.     Ans. 


TABLE  I. 


SPECIFIC    GRAVITY   AND   WEIGHT   PER   CUBIC   FOOT   OF 
VARIOUS  METALS. 


Substance. 

Specific 
Gravity. 

Weight  per 
Cubic  Foot. 
Pounds. 

Platinum  

21.50 

1  343  8 

Gold  

19.50 

1  218.8 

Mercury 

13  60 

850  0 

Lead  (ca'st)          

11.35 

709.4 

Silver  

10.50 

656.3 

Copper  (cast) 

8  79 

549  4 

Brass                 

8.38 

523  8 

Wrought  Iron  

7.68 

480.0 

Cast  Iron 

7  21 

450  0 

Steel 

7.84 

490  0 

Tin  (cast)  

7.29 

455.6 

Zinc  (cast)  

6.86 

428  8 

Antimony 

6  71 

419  4 

Aluminum 

2.50 

156*.  3 

10 


MECHANICS  OF  FLUIDS. 
TABLE  n. 


SPECIFIC    GRAVITY    AND    WEIGHT    PER    CUBIC    FOOT    OF 
VARIOUS  WOODS. 


Substance. 

Specific 
Gravity. 

Weight  per 
Cubic  Foot. 
Pounds. 

Ash 

845 

52  80 

Beech      

852 

53  .  25 

Cedar 

561 

35  06 

Cork 

240 

15  00 

Ebony  (American)  

1  331 

83.19 

Lignum-  vitas 

1  333 

83  30 

Maple  

750 

46.88 

Oak  (old) 

1  170 

73  10 

Spruce 

500 

31  25 

Pine  (yellow)     ... 

.060 

41.20 

Pine  (white)  

.554 

34.60 

Walnut 

671 

41  90 

TABLE  III. 


SPECIFIC    GRAVITY    AND    WEIGHT    PER    CUBIC    FOOT    OF 
VARIOUS   LIQUIDS. 


Substance. 

Specific 
Gravity. 

Weight  per 
Cubic  Foot. 
Pounds. 

Acetic  acid 

1  062 

66  4 

Nitric  acid  

1  217 

76  1 

Sulphuric  acid  
Muriatic  acid 

1.841 
1  200 

115.1 

75  0 

Alcohol   

.800 

50.0 

Turpentine  

.870 

54.4 

Sea  water  (ordinary) 

1  026 

64  1 

Milk       

1.032 

64.5 

MECHANICS  OF.  FLUIDS. 


17 


25.  The  specific  gravities  of  different  bodies  are  given  in 
printed  tables;  hence,  if  it  is  desired  to  know  the  weight  of 
a  body  that  cannot  be  conveniently  weighed,  calculate  its 
cubical  contents  and  multiply  the  specific  gravity  of  the  body 
by  the  weight  of  a  like  volume  of  water,  remembering  that  a 
cubic  foot  of  water  weighs  62.^2  pounds. 

EXAMPLE  1. — How  much  will  3,214  cubic  inches  of  cast  iron  weigh  ? 
Take  its  specific  gravity  as  7.21. 

SOLUTION. — Since  1  cubic  foot  of  water  weighs  62.42  pounds,  3,214 
cubic  inches  weigh 


3,214 


X  62.42  =  116.098  pounds. 


Then, 


1,728 
116.098  X  7.21  =  837.067  Ib.     Ans. 


TABLE  IV. 


SPECIFIC   GRAVITY   AND   WEIGHT   PER   CTTBIC   FOOT    OF 

VARIOUS    GASES   AT   3S8f   F.    AND   UNDER   A 

PRESSURE    OF    1    ATMOSPHERE. 


Substance. 

Specific 
Gravity. 

Weight  per 
Cubic  Foot. 
Pounds. 

Atmospheric  air  

1.0000 

08073 

Carbonic  acid 

1  5290 

12344 

Carbonic  oxide 

9674 

07810 

Chlorine             '    

2  .  4400 

19700 

Oxygen  

1.1056 

.08925 

Nitrogen 

9736 

07860 

Smoke  (bituminous  coal)  

.1020 

00815 

Smoke  (wood)  

.0900 

.00727 

*Steam  at  212°  F 

4700 

03790 

Hydrogen                

.0692 

.00559 

*  The  specific  gravity  of  steam  at  any  temperature  and  pressure 
compared  with  air  at  the  same  temperature  and  pressure  is  .622. 


18  MECHANICS  OF  FLUIDS.  §  6 

EXAMPLE  2.— What  is  the  weight  of  a  cubic  inch  of  cast  iron  ? 
SOLUTION.—    ^||  X  7.21  =  .26044  Ib.     Ans. 

One  cubic  foot  of  pure  distilled  water  at  a  temperature  of 
39.2°  Fahrenheit  weighs  62.425  pounds,  but  t lie  value  usually 
taken  in  making  calculations  is  62.5  pounds. 

EXAMPLE  3. — What  is  the  weight  in  pounds  of  7  cubic  feet  of 
oxygen  ? 

SOLUTION. — One  cubic  foot  of  air  weighs  .08073  pound,  and  the 
specific  gravity  of  oxygen  is  1.1056,  compared  with  air;  hence, 

.08073  X  1.1056  X  7  =  .62479  Ib.,  nearly.     Ans. 


TABLE  V. 


SPECIFIC   GRAVITY   AND   WEIGHT   PER   CUBIC   FOOT   OF 
VARIOUS   SUBSTANCES. 


Substance. 

Specific 
Gravity. 

Weight  per 
Cubic  Foot. 
Pounds. 

Emery   .      .      .    .           

4.00 

250 

Glass  (average)  
Chalk 

2.80 
2  78 

175 
174 

Granite 

2.65 

166 

Marble  

2.70 

169 

Stone  (common)  

2.52 

158 

Salt  (common) 

2.13 

133 

Soil  (common)  

1.98 

124 

Clay 

1.93 

121 

Brick                                                  .... 

1.90 

118 

Plaster  of  Paris  (average)  

2.00 

125 

Sand                                               .  . 

1.80 

113 

MECHANICS  OF  FLUIDS. 


BUOYANT  EFFECTS  OF  WATER. 

26.  In  Fig.  11  is  shown  a  6-inch  cube  entirely  submerged 
in  water.  The  lateral  pressures  are  equal  and  in  opposite 
directions.  The  upward  pressure  acting  on 
the  lower  surface  of  the  cube  is  6  X  6  X  21 
X  .03617 ;  the  downward  pressure  acting  on 
the  top  of  the  cube  is  6  X  6  X  15  X  .03617; 
and  the  difference  is  6  X  6  X  6  X  .03617, 
which  equals  the  volume  of  the  cube  in 
cubic  inches  times  the  weight  of  1  cubic 
inch  of  water.  That  is,  the  upward  pres- 
sure exceeds  the  downward  pressure  by 
the  weight  of  a  volume  of  water  equal  to 
the  volume  of  the  body. 

This  excess  of  upward  pressure  over  the 

downward  pressure  acts  against  gravity ;  that  is,  the  water 
presses  the  body  upwards  with  a  greater  force  than  it  presses 
it  downwards ;  consequently,  if  a  body  is  immersed  in  a 
fluid,  it  ^vill  lose  in  weight  an  amount  equal  to  the  weight  of 
the  fluid  it  displaces.  This  is  called  the  principle  of  Archi- 
medes, because  it  was 
first  stated  by  him. 

27.  This  principle 
may  be  experimentally 
demonstrated  with  the 
beam  scales,  as  shown  in 
Fig.  12.  From  one  scale 
pan  suspend  a  hollow 
cylinder  of  metal  t,  and 
below  that  a  solid  cylin- 
der a  of  the  same  size  as 
the  hollow  part  of  the 
upper  cylinder.  Put 
weights  in  the  other 
scale  pan  until  they  ex- 
actly balance  the  two 
in  water,  the  scale  pan 


FIG.  12. 


cylinders.      If    a    be    immersed 


20  MECHANICS  OF  FLUIDS.  §  6 

containing  the  weights  will  descend,  showing  that  a  has 
lost  some  of  its  weight.  Now  fill  t  with  water,  and  the 
volume  of  water  that  can  be  poured  into  t  will  equal  that 
displaced  by  a.  The  scale  pan  that  contains  the  weights 
will  gradually  rise  until  t  is  filled,  when  the  scales  again 
balance. 

If  a  body  be  lighter  than  the  liquid  in  which  it  is  immersed, 
the  upward  pressure  will  cause  it  to  rise  and  project  partly 
out  of  the  liquid,  until  the  weight  of  the  body  and  the  weight 
of  the  liquid  displaced  are  equal.  If  the  immersed  body  be 
heavier  than  the  liquid,  the  downward  pressure  plus  the 
weight  of  the  body  will  be  greater  than  the  upward  pressure, 
and  the  body  will  fall  downwards  until  it  touches  bottom  or 
meets  an  obstruction.  If  the  weights  of  equal  volumes  of 
the  liquid  and  the  body  are  equal,  the  body  will  remain 
stationary  and  will  be  in  equilibrium  in  any  position  or 
depth  beneath  the  surface  of  the  liquid. 

28.  An  interesting   experiment    in  confirmation  of   the 
above  facts  may  be  performed  as  follows :     Drop  an  egg  into 
a  glass  jar  filled  with  fresh  water.     The  mean  density  of  the 
egg  being  a  little  greater  than  that  of  the  water,  it  will  fall 
to  the  bottom  of  the  jar.      Now,  dissolve  salt  in  the  water, 
stirring  it  so  as  to  mix  the  fresh  and  salt  water.     The  salt 
water  will  presently  become  denser  than  the  egg  and  the 
egg  will  rise.     Now,  if  fresh  water  be  poured  in  until  the 
egg  and  water  have  the  same  density,  the  egg  will  remain 
stationary  in  any  position  that  it  may  be  placed  below  the 
surface  of  the  water. 

29.  The  principle  of  Archimedes  gives  a  very  easy  and 
accurate  method  of  finding  the   volume  of   an   irregularly 
shaped  body.     Thus,  subtract  its  weight  in  water  from  its 
weight  in  air    and  divide  by  .03617;   the  quotient  will  be 
the  volume  in  cubic  inches,  or  divide  by  62.5  and  the  quo- 
tient will  be  the  volume  in  cubic  feet. 

If  the  specific  gravity  of  the  body  is  known,  its  cubical 
contents  can  be  found  by  dividing  its  weight  by  its  specific 
gravity,  and  then  dividing  again  by  either  .03617  or  62.5. 


MECHANICS  OF  FLUIDS. 


EXAMPLE. — A  certain  body  has  a  specific  gravity  of  4.38  and  weighs 
76  pounds;  how  many  cubic  inches  are  there  in  the  body  ? 
76 


SOLUTION. — 


4. 38  X. 03617 


=  479.72  cu.  in.     Ans. 


THE  HYDROMETER. 

30.  Instruments  called  hydrometers  are  in  general  use 
for  determining  quickly  and  accurately  the  specific  gravities 
of  liquids  and  some  forms  of  solids.     They  are  of  two  kinds, 
viz. :    (1)   Hydrometers  of  constant  weigJit,   as   Beaume's ; 
(2)  hydrometers  of  constant  volume,  as  Nicholson's. 

A  hydrometer  of  constant  weight  is  shown  in  Fig.  13. 
It  consists  of  a  glass  tube,  near  the  bottom  of  which  are  two 
bulbs.  The  lower  and  smaller  bulb  is  loaded 
with  mercury  or  shot,  so  as  to  cause  the 
instrument  to  remain  in  a  vertical  position 
when  placed  in  the  liquid.  The  upper  bulb  is 
filled  with  air,  and  its  volume  is  such  that  the 
whole  instrument  is  lighter  than  an  equal 
volume  of  water. 

The  point  to  which  the  hydrometer  sinks 
when  placed  in  water  is  usually  marked,  the 
tube  being  graduated  above  and  below  in  such 
a  manner  that  the  specific  gravity  of  the  liquid 
can  be  read  directly.  It  is  customary  to  have 
two  instruments,  one  with  the  zero  point  near 
the  top  of  the  stem  for  use  in  liquids  heavier 
than  water,  and  the  other  with  the  zero  point  near  the  bulb 
for  use  in  liquids  lighter  than  water. 

These  instruments  are  more  commonly  used  for  determin- 
ing the  degree  of  concentration  or  dilution  of  certain  liquids, 
as  acids,  alcohol,  milk,  solutions  of  sugar,  etc.,  rather  than 
their  actual  specific  gravities.  They  are  then  known  as 
acidometers,  alcoholometers,  lactometers,  saccharometers,  sali- 
nometers,  etc. ,  according  to  the  use  to  which  they  are  put. 

31.  Hydrometers  of  constant  volume  are  not  in  com- 
mon use  and  are  rarely  found  outside  of  laboratories. 


22  MECHANICS  OF  FLUIDS. 

HYDROKI^TETICS. 


FLOW  OF  WATER  IX  PIPES. 

32.  Experience  has  demonstrated  that  for  satisfactory 
work,  the  flow  of  water  in  the  suction  pipes  of  boiler  feed- 
pumps  and   other  comparatively  small   pumps  should    not 
exceed  200  feet  per  minute,  and  it  should  not  be  more  than 
500  feet  in  the  delivery  pipe  for  a  duplex  pump,  or  400  feet 
for  a  single-cylinder  pump. 

Knowing  the  volume  of  water  that  is  to  flow  through  or 
to  be  discharged  from  a  pipe  in  1  minute,  the  area  of  the 
suction  and  delivery  pipes  can  readily  be  determined. 

33.  The  volume  of  water  in  cubic  feet  discharged  from 
a  pipe  in  1  minute  is  equal  to  the  velocity  in  feet  per  minute 
times  the  area  of  the  pipe  in  square  feet.     Then,  the  area  of 

,    volume  in  cubic  feet  per  minute        . 

the  pipe  equals —        —. : — -. —          — ; —      — .     As  there 

velocity  in  feet  per  minute 

are  144  square  inches  in  a  square  foot,  the  area  of  the  pipe 

.    144  X  volume  in  cubic  feet  per  minute 
in  square  inches  is  —  — . 

velocity  in  feet  per  minute 

34.  If  the  volume  is  expressed  in  gallons  per  minute, 
then,  as  there  are  7.48  gallons  in  a  cubic  foot,  the  area  of  the 

144  X  volume  in  gallons* 

pipe  in  square  inches  will  be —  — = . 

7.48  X  velocity  in  feet  per  minute 

Hence,  the  following  rule, 

where  n  =  gallons  per  minute ; 

v  =  velocity  in  feet  per  minute ; 
A  =  area  of  pipe  in  square  inches. 

Rule  4. —  To  find  the  area  of  a  pipe  in  square  inches, 
divide  19.25  times  the  number  of  gallons  per  minute  by  the 
velocity  in  feet  per  minute. 

19.25  « 


Or,  A  = 


v 


*  The  gallon  here  referred  to  is  the  Winchester  or  wine  gallon 
of  231  cubic  inches  capacity  and  in  common  use  in  the  United  States 
of  America. 


§  6  MECHANICS  OF  FLUIDS.  23 

EXAMPLE.  —  If  a  duplex  pump  is  used,  what  area  of  feed-pipe  is 
required  for  a  boiler  into  which  25  gallons  of  water  per  minute  is  to 
be  pumped  ? 

SOLUTION.—  The  allowable  velocity  being  500  feet,  by  applying 
rule  4,  we  get 

19.25  X  25 
=  -       —  =  -9625  s-  ln-     Ans- 


35.  The  quantity  of  water,  expressed  in  gallons,  that 
will  flow  through  a  given  pipe  in  1  minute,  its  velocity  being 
known,  is  given  by  the  following  rule: 

Rule  5.  —  Multiply  the  area  of  the  pipe  in  square  inches  by 
the  velocity  in  feet  per  minute.  Divide  the  product  by  19.  25. 

Av 


EXAMPLE.  —  How  many  gallons  of  water  per  minute  will  flow 
through  a  pipe  having  an  area  of  2  square  inches,  the  velocity  of 
flow  being  450  feet  per  minute  ? 

SOLUTION.  —  Applying  the  rule  just  given,  we  get 

Ans- 

36.  The  velocity  with  which  water  will  flow  through  the 
delivery  pipe  of  a  pump  when  the  area  of  the  water  cylinder, 
the  area  of  the  delivery  pipe,  and  the  piston  speed  of  the 
pump  are  known,  is  given  by  the  following  rule, 

where          v  =  velocity  in  feet  per  minute  ; 

A  =  area  of  delivery  pipe  in  square  inches; 
a  =  area  of  water  piston  in  square  inches; 
S  =  piston  speed  in  feet  per  minute. 

Rule  6.  —  Multiply  the  area  of  the  water  piston  by  the 
piston  speed,  and  divide  this  product  by  the  area  of  the  delivery 
pipe. 

aS 
Or,  p.p-p 

EXAMPLE.  —  If  the  water  piston  of  a  pump  has  an  area  of  12  square 
inches  and  moves  at  a  speed  of  100  feet  per  minute,  what  will  be  the 

H.  S.    I.—  19 


24  MECHANICS  OF  FLUIDS.  §  6 

velocity  of  the  water  in  the  delivery  pipe  if  the  latter  has  an  area 
of  2  square  inches  ? 

SOLUTION.— Applying  rule  6,  we  get 

v  =  12xl°°  =  600  ft.  per  min.     Ans. 


STANDARD  PIPE  DIMENSIONS. 

37.  The  great  majority  of  the  pipes  used  about  steam 
plants  are  made  of  wrought  iron  and  are  almost  invariably 
made   in  accordance  with  the  Briggs  standard.      It  will  be 
noticed  that  the  diameter  of  the  pipe  by  which  it  is  known 
to  the  trade  is  not  the  actual  diameter;  hence,  in  calcula- 
ting the  amount  of  water  that  will  flow  through  a  pipe,  the 
actual  diameter  or  actual  internal  area  must  be  taken  from 
Table  VI. 

38.  As  wrought-iron  pipes  are  not  made  in  sizes  differing 
from  those  given  in  the  table,  it  will  be  apparent  that  only 
in  rare  instances  can  a  pipe  be  selected  that  will  have  the 
area  calculated  by  rule  4.     In  practice  the  nearest  commer- 
cial size  of  pipe  would  be  selected. 

EXAMPLE. — What  commercial  size  of  delivery  pipe  should  be  used 
for  a  single-cylinder  pump  to  deliver  90  gallons  of  water  per  minute  ? 

SOLUTION. — For  a  single-cylinder  pump,  the  velocity  of  flow  should 
not  be  more  than  400  feet  per  minute.  Then,  applying  rule  4,  we  get 

19.25  X  90 
= 400 =  4-33  scl-  in- 

According  to  Table  VI,  the  commercial  size  of  pipe  having  an  area 
nearest  this  is  2£  inches;  therefore,  a  2^-inch  pipe  should  be  used. 

Ans. 

Rule  4  will  be  found  to  agree  quite  closely  with  the  prac- 
tice of  the  leading  pump  manufacturers.  In  case  they 
should,  however,  recommend  a  larger  size  of  pipe,  it  is 
advisable  to  follow  their  advice. 

Pipes  made  in  accordance  with  the  Briggs  standard,  when 
below  1  inch  nominal  size,  are  butt-welded  and  proved  to 
300  pounds  per  square  inch  by  hydraulic  pressure.  Pipes 
above  1  inch  are  lap-welded  and  proved  to  500  pounds. 


MECHANICS  OF  FLUIDS. 
TABLE    VI. 


TABLE  OF  STANDARD  DIMENSIONS  OF  WROUGHT-IRON 
WELDED  PIPES. 


Nominal 
Diameter. 
Inches. 

Actual 
Internal 
Diameter. 
Inches. 

Actual 
Internal 
Area. 
Square 
Inches. 

Actual 
External 
Diameter. 
Inches. 

Number  of 
Threads 
Per  Inch. 

* 

.27 

.057 

.40 

27 

i 

.36 

.104 

.54 

18 

t 

.49 

.192 

.67 

18 

1 

.62 

.305 

.84 

14 

t 

.82 

.533 

1.05 

14 

1 

1.05 

.863 

1.31 

Hi 

H 

1.38 

1.496 

1.66 

Hi 

ii 

1.61 

2  .  038 

1.90 

11* 

2 

2.07 

3.355 

2.37 

Hi 

n 

2.47 

4.783 

2.87 

8 

3 

3.07 

7.388 

3.50 

8 

3i 

3.55 

9.887 

4.00 

8 

4 

4.07 

12.730 

4.50 

8 

4* 

4.51 

15.939 

5.00 

8 

5 

5.04 

19  990 

5.56 

8 

6 

6.06 

28.889 

6.62 

8 

7 

7.02 

38.737 

7.62 

8 

8 

7.98 

50  .  039 

8.62 

8 

9 

9.00 

63  .  633 

9.62 

8 

10 

10  .  02 

78.838 

10.75 

8 

PIPE  FITTINGS. 

39.  In  piping  a  steam  plant,  it  is  rarely  possible  to  run 
the  piping  in  a  straight  line,  it  being  usually  necessary  to 
introduce  one  or  more  elbows  or  similar  fittings  to  reach  the 
point  desired.  The  effect  of  T  and  L  fittings  is  to  increase 
the  resistance  to  the  flow  of  the  water  through  the  pipes, 


26  MECHANICS  OF  FLUIDS.  §  6 

thus  requiring  the  pump  to  do  more  work  for  the  same  quan- 
tity of  water  delivered. 

As  pumps  for  boiler  feeding  are  always  built  with  a  large 
steam  cylinder,  there  is  enough  excess  of  power  to  allow  the 
pump  under  ordinary  conditions  to  force  the  required  quan- 
tity of  water  through  the  pipe.  On  the  suction  side  of  the 
pump,  however,  the  force  impelling  the  water  to  flow  into 
the  pump  is  quite  small,  and  a  very  slight  resistance  will  be 
sufficient  to  interfere  with  the  flow  of  the  water  into  the 
pump.  Hence,  it  is  important  that  the  suction  pipe  be  as 
straight  as  possible ;  if  it  is  impossible  to  make  it  straight, 
larger  sizes  of  pipe  should  be  used,  or  easy  bends  with  as 
large  a  radius  as  possible  should  be  substituted  for  the 
elbows.  This  applies  especially  to  pumps  that  must  lift  the 
water  more  than  10  feet.  It  is  impossible  to  lay  down  any 
hard-and-fast  rules  as  to  what  numbers  of  elbows  should  not 
be  exceeded  in  a  suction  and  delivery  pipe ;  judgment  will 
have  to  be  used.  Generally  speaking,  they  should  be  as 
few  as  possible. 

EXAMPLES  FOR  PRACTICE. 

1.  Suppose  a  cylinder  to  be  filled  with  water  and  placed  in  an 
upright  position.     If  the  diameter  of  the  cylinder  is  19  inches  and  its 
total  length  inside  26  inches,  what  will  be  the  total  pressure  on  the 
bottom  when  a  pipe  £  inch  in  diameter  and  12  feet  long  is  screwed 
into  the  cylinder  head  and  filled  with  water  ?     The  pipe  is  vertical. 

Ans.  1,743.2  Ib. 

2.  In  the  last  example,  what  is  the  total  pressure  against  the  upper 
head?  Ans.  1,476.6  Ib. 

3.  In  example  1,  a  piston  is  fitted  to  the  upper  end  of  the  pipe  and 
an  additional  force. of  10  pounds  is  applied  to  the  water  in  the  pipe. 
What  is  the  total  pressure  (a)  on  the  bottom  of  the  cylinder  ?    (b)  on 
the  upper  head  ?  j  (a)     16,184  Ib. 

Lns'  1  (b)     15,906  Ib. 

4.  In  example  3,  what  is  the  pressure  per  square  inch  in  the  pipe 

2  inches  from  the  upper  cylinder  head  ?          Ans.  56.0656  Ib.  per  sq.  in. 

5.  A  water  tower  80  feet  high  is  filled  with  water.     A  pipe  4  inches 
in  diameter  is  so  connected  to  the  side  of  the  tower  that  its  center  is 

3  feet  from  the  bottom.     If  the  pipe  is  closed  by  a  flat  cover,  what  is 
the  total  pressure  against  the  cover  ?  Ans.  420  Ib. 


MECHANICS  OF  FLUIDS. 


27 


6.  In  the  last  example,  what  is  the  upward  pressure  per  square  inch 
10  feet  from  the  bottom  of  the  tower  ?  Ans.  30.3828  Ib.  per  sq.  in. 

7.  A  cube  of  wood,   one  edge  of  which  measures  3  feet,  is  sunk 
until  the  upper  surface  is  40  feet  below  the  level  of  the  water ;  what  is 
the  total  force  that  tends  to  move  the  cube  upwards  ?    Ans.  1,687.5  Ib. 

8.  A  body  weighs  72  pounds  when  immersed  in  water  and  321  pounds 
in  air;  what  is  its  volume  in  cubic  feet  ?  Ans.  3.984  cu.  ft. 

9.  The  specific  gravity  of  a  body  being  9.5  and  its  weight  81  pounds, 
what  is  its  volume  in  cubic  inches  ?  Ans.  235.73  cu.  in. 

10.  What  commercial  size  of  pipe  should  be  used  for  the  suction 
and  delivery  pipes  of  a  single-cylinder  pump  to  deliver  50  gallons  of 

water  per  minute  ?  .         (  Suction  pipe,  2^  in. 

Ans.  •{,-..,.  .        ~  . 

(  Delivery  pipe,  2  in. 

11.  How  many  gallons  per  hour,  at  a   velocity  of  400  feet  per 
minute,  will  flow  through  a  1-inch  pipe?  Ans.  1,076  gal.,  nearly. 

12.  The  water  piston  of  a  pump  is  3£  inches ;  the  piston  speed  is 
54  feet  per  minute;  and  a  l|-inch  delivery  pipe  is  used;  what  is  the 
velocity  in  the  delivery  pipe  ?  Ans.  255  ft.  per  min.,  nearly. 


PNEUMATICS. 


PRESSURE  OF  GASES. 

40.  Pneumatics  is  that  branch  of  mechanics  that  treats 
of  the  properties  and  pressures  of  gases. 

41.  The  most  striking  feature  of  all  gases  is  their  great 
expansibility.     If  we  inject  a 

quantity  of  gas,  ho^vever  small, 
into  a  vessel,  it  will  expand 
and  fill  that  vessel.  If  a  blad- 
der or  football  be  partly  filled 
with  air  and  placed  under  a 
glass  jar  (called  a  receiver), 
from  which  the  air  has  been 
exhausted,  the  bladder  or  foot- 
ball will  immediately  expand, 
as  shown  in  Fig.  14.  The 
force  that  a  gas  always  exerts 
when  confined  in  a  limited 


28  MECHANICS  OF  FLUIDS.  §  6 

space  is  called  tension.     The  word  tension  in  this  case  means 
pressure,  and  is  only  used  in  this  sense  in  reference  to  gases. 

4:2.  As  water  is  the  most  common  type  of  fluids,  so  air 
is  the  most  common  type  of  gases.  It  was  supposed  by  the 
ancients  that  air  had  no  weight,  and  it  was  not  until  about 
the  year  1650  that  the  contrary  was  proved.  A  cubic  inch 
of  air,  under  ordinary  conditions,  weighs  .31  grain,  nearly. 
At  a  temperature  of  32°  F.  and  a  pressure  of  14.7  pounds 
per  square  inch,  the  ratio  of  the  weight  of  air  to  water  is 
about  1  :  774 ;  that  is,  air  is  only  -^^  as  heavy  as  water.  It 
has  been  shown  that  if  a  body  were  immersed  in  water  and 
weighed  less  than  the  volume  of  water  displaced,  the  body 
would  rise  and  project  partly  out  of  the  water.  The  same 
is  true,  to  a  certain  extent,  of  air.  If  a  vessel  made  of  light 
material  is  filled  with  a  gas  lighter  than  air,  so  that  the 
total  weight  of  the  vessel  and  gas  is  less  than  the  air  they 
displace,  the  vessel  will  rise.  It  is  on  this  principle  that 
balloons  are  made. 


PRESSURE  OF  THE   ATMOSPHERE. 

43.  Since  air  has  weight,  it  is  evident  that  the  enormous 
quantity  of  air  that  constitutes  the  atmosphere  must  exert 
a  considerable  pressure  on  the  earth.  This  is  easily  proved 
by  taking  a  long  glass  tube  closed  at  one  end  and  filling  it 
with  mercury.  If  the  finger  be  placed  over  the  open  end  so 
as  to  keep  the  mercury  from  running  out  and  the  tube 
inverted  and  placed  in  a  cup  of  mercury,  as  shown  in  Fig.  15, 
the  mercury  will  fall,  then  rise,  and  after  a  few  oscillations 
will  come  to  rest  at  a  height  above  the  top  of  the  mercury 
in  the  cup  equal  to  about  30  inches  at  sea  level. 

This  height  will  always  be  the  same  under  the  same 
atmospheric  conditions.  Now,  if  the  atmosphere  has 
weight,  it  must  press  upon  the  upper  surface  of  the  mercury 
in  the  cup  with  equal  intensity  upon  every  square  unit, 
except  upon  that  part  of  the  surface  occupied  by  the  tube. 
In  order  that  there  may  be  equilibrium,  the  weight  of  the 
mercury  in  the  tube  must  be  equal  to  the  pressure  of  the  air 


MECHANICS  OF  FLUIDS. 


29 


upon  a  portion  of  the  surface  of  the  mercury  in  the  cup 
equal  in  area  to  the  inside  of  the  tube.  Suppose  that  the 
area  of  the  inside  of  the  tube  is  1  square  inch,  then,  since 
mercury  is  13.6  times  as  heavy  as  water,  the  weight  of  the 
mercurial  column  is  .03617  X  13.6  X  30  =  14.7574  pounds. 
The  actual  height  of  the  mercury 
is  a  little  less  than  30  inches,  and 
the  actual  weight  of  a  cubic  inch 
of  distilled  water  is  a  little  less 
than  .03617  pound.  When  these 
considerations  are  taken  into  ac- 
count, the  average  weight  of  the 
mercurial  column  at  the  level  of 
the  sea,  when  the  temperature  is 
60°  F.,  is  14.69  pounds,  or  prac- 
tically 14.7  pounds.  Since  this 
weight,  when  exerted  upon  1  square 
inch  of  the  liquid  in  the  glass,  just 
produces  equilibrium,  it  is  plain 
that  the  pressure  of  the  outside 
air  is  14. 7  pounds  upon  every  square 
inch  of  surface. 

44.  Vacuum. — The    space    be- 
tween the  upper  end  of  the  tube 
and  the  upper  surface  of  the  mer- 
cury is  called  a  vacuum,  meaning 
that  it  is  an  entirely  empty  space 
and    does    not    contain    any    sub- 
stance— solid,    liquid,    or    gaseous.  FIG-  15- 

If  there  were  a  gas  of  some  kind  there,  no  matter  how  small 
the  quantity  might  be,  it  would  expand  and  fill  the  space, 
and  its  tension  would  cause  the  column  of  mercury  to  fall 
and  become  shorter,  according  to  the  amount  of  gas  or  air 
present.  The  condition  then  existing  in  the  space  would  be 
called  a  partial  vacuum. 

45.  The  Measurement  of  Vacuum. — The   degree    to 
which  the  air  has  been  exhausted  from  a  closed  vessel  in 


30 


MECHANICS  OF  FLUIDS. 


(i 


which  there  is  a  partial  vacuum  is  measured  by  the  height 
to  which  a  mercurial  column  in  a  vertical  tube,  whose  top  is 
connected  to  the  vessel,  will  rise  under  the  pressure  of  the 
atmosphere.  Thus,  when  we  say  that  there  is  a  vacuum  of 
29  inches  in  a  vessel,  we  mean  that  enough  air  has  been 
exhausted  from  the  vessel  to  enable  the  surrounding  air  to 
support  a  mercurial  column  29  inches 
high,  as  shown  in  Fig.  16,  where  A  is 
the  vessel  in  which  there  is  a  partial 
vacuum.  In  other  words,  the  pressure 
in  the  vessel  is  less  than  the  pressure 
of  the  atmosphere  by  a  column  of  mer- 
cury 29  inches  in  height. 

When  there  is  a  partial  vacuum  in- 
side a  closed  vessel,  the  air  remaining 
in  the  vessel  is  under  a  tension,  or  pres- 
sure, less  than  its  original  tension  of 
14.7  pounds  per  square  inch.  This  ten- 
sion, in  inches  of  mercury,  is  equal  to 
the  difference  between  the  height  at 
which  the  mercurial  column  connected 
to  the  vessel  stands  and  that  at  which 
it  would  stand  if  the  vacuum  were  per- 
fect. Consider  that  the  mercury  col- 
umn will  be  in  equilibrium  when  the 
hydrostatic  pressure  on  its  base  equals 
the  atmospheric  pressure.  The  hydro- 
static pressure  on  the  base  is  the  sum 
of  two  pressures:  (1)  The  pressure  due 
to  the  weight  of  the  mercury  column; 
(2)  the  pressure  in  the  space  above  the 
mercury. 

From  this  it  follows  that  if  the  atmospheric  pressure  and 
the  pressure  due  to  the  weight  of  the  mercurial  column  are 
given,  the  pressure  above  the  column  must  be  that  due  to 
their  difference.  As  the  atmosphere  will  force  a  mercurial 
column  to  a  height  of  30  inches  when  there  is  a  perfect  vac- 
uum above  it,  it  follows  that  to  find  the  pressure  in  a  vessel 


FIG.  16. 


§  6  MECHANICS  OF  FLUIDS.  31 

in  which  there  is  a  partial  vacuum,  the  number  of  inches  of 
height  of  the  mercurial  column  is  to  be  subtracted  from  30. 
Thus,  if  the  vacuum  in  a  vessel  is  26  inches,  the  pressure  in 
the  vessel  is  30  —  26  =  4  inches  of  mercury. 

46.  In  practice  it  is  convenient  to  always  use  the  same 
unit  of  pressure,  which  is  1  pound  per  square  inch.     We 
know  that  14.7  pounds  per  square  inch  will  support  a  mercu- 
rial column  30  inches  high.      Hence,  1  inch  of  height  of  a 

14  7 

mercurial  column  represents  a  pressure  of  —  '—  =  .49  pound 

oO 

per  square  inch.  Then,  to  reduce  inches  of  mercury  to 
pounds  per  square  inch,  multiply  their  number  by  .49. 

EXAMPLE.— A  gauge  shows  a  vacuum  of  22  inches  in  a  condenser. 
What  is  the  absolute  pressure  in  the  condenser  ? 

SOLUTION. — The  pressure,  in  inches  of  mercury,  is  30  —  22  =  8  inches, 
or  8  X  .49  =  3.92  Ib.  per  sq.  in.  Ans. 

47.  The  Vacuum  Gauge. — For  engineering  work,  the 
glass  tube  of  Fig.  16  would  be  a  rather  inconvenient  form  of 
gauge  for  measuring  the  vacuum.     Hence,  special  metallic 
gauges,    known  as  vacuum  gauges,  have  been  designed. 
Their  dial  is  graduated  to  show  the  degree  of  vacuum,  in 
inches  of  mercury.     In  steam-engineering  work,   they  are 
used  chiefly  in  connection  with  condensers  for  steam  engines. 
It  should  be  borne  in  mind  that  a  vacuum  gauge  does  not 
indicate  the  pressure  in  the  vessel  to  which  it  is  attached, 
but  instead  shows,   in  inches  of  mercury,  how  much   the 
pressure  has  been  lowered  below  the  atmospheric  pressure. 
In  this  respect  a  vacuum  gauge  differs  essentially  from  a 
pressure  gauge,  which  shows  how  much  the  pressure  has 
been  increased  either  above  the  pressure  of  a  perfect  vac- 
uum, which  is  zero,  as  in  case  of  a  gauge  registering  the 
pressure  of  the  atmosphere,  or  above  the  atmospheric  pres- 
sure, as  in  case  of  the  ordinary  pressure  gauge. 

48.  If  the  tube  of  Fig.  15  had  been  filled  with  a  liquid 
lighter  than  mercury,  the  height  of  the  column  required  to 
balance  the  atmospheric  pressure  would  be  greater.     The 


MECHANICS  OF  FLUIDS. 


height  of  the  column  depends  on  the  specific  gravity  of  the 
liquid.     Thus,  if  the  liquid  had  a  specific  gravity  of  1,  as 

water,  its  height  would  be  —  —  =  408  inches,  or  34  feet. 

This  means  that  if  a  tube  be  filled  with  water, 
inverted,  and  placed  in  a  dish  of  water,  in  a 
manner  similar  to  the  experiment  made  with  the 
mercury,  the  height  of  the  column  of  water  will 
be  34  feet. 

49.  The  Barometer. — As  is  well  known,  the 
atmosphere   does    not   exert   exactly   the   same 
pressure  at  all  times;  the  pressure  varies  with 
conditions.     As  the  height  of   the   mercury  in 
the  glass  tube  of   Fig.   15   depends   chiefly  on 
the  atmospheric  pressure,  it  is  evident  that  an 
instrument   constructed   on   this   principle   will 
indicate  the  varying  pressure  by  the  height  of 
the   column.      Such   an   instrument  is  called  a 
mercurial  barometer. 

50.  A  standard  form  of  barometer  is  shown 
in  Fig.  17.     The  barometer  is  simply  a  pressure 
gauge  that  registers  the  pressure  of  the  air.     In 
this  case  the  cup  and  tube  at  the  bottom  are  pro- 
tected by  a  brass  or  iron  casing.     At  the  top  of 
the  tube  is  a  graduated  scale.     Attached  to  the 
casing  is  an  accurate  thermometer  for  determin- 
ing the   temperature  of  the  outside  air  at  the 
time  the  barometric  observation  is  taken.      This 
is  necessary,   since  mercury  expands  when  the 
temperature    is   increased    and   contracts  when 
the  temperature  falls ;  for  this  reason  a  standard 
temperature  is  assumed,  and  all  barometer  read- 
ings   are    reduced    to    this    temperature.     This 
standard  temperature  is  usually  taken  at  32°  F., 

at  which  temperature  the  height  of  the  mercurial  column 
is  30  inches  under  normal  conditions  at  sea  level.     Another 


§  6  MECHANICS  OF  FLUIDS.  33 

correction  is  made  for  the  altitude  of  the  place  above  sea 
level,  and  a  third  correction  for  the  effects  of  capillary 
attraction. 

51.  A  mercurial  barometer  is  not  only  a  very  bulky 
instrument,  but  is  also  quite  heavy  and,  hence,  is  not  very 
well  adapted  for  transportation.  To  overcome  these  draw- 
backs, a  form  of  portable  barometer  known  as  an  aneroid 
barometer,  has  been  designed,  which  operates  on  a  some- 
what different  principle.  Such  a  barometer  is  shown  in 


FIG.  18. 

Fig.  18.  The  principal  part  of  the  aneroid  barometer  is  a 
cylindrical,  air-tight  box  of  metal  closed  by  a  corrugated  top 
of  thin,  elastic  metal.  The  air  is  exhausted  from  the  box, 
which  is  then  sealed.  Evidently,  the  pressure  of  the  air  on 
the  outside  of  the  cover  will  cause  the  cover  to  curve  inwards, 
as  the  space  inside  of  the  cover  is  void  of  pressure,  until  the 


34  MECHANICS  OF  FLUIDS.  §  6 

resistance  due  to  the  elasticity  of  the  cover,  aided  by  the 
resistance  of  a  spring  beneath  it,  is  equal  to  the  force  exerted 
by  the  air.  Now,  if  the  air  pressure  increases,  the  cover 
will  be  curved  further  inwards ;  if  the  air  pressure  decreases, 
the  elasticity  of  the  cover,  aided  by  the  spring  beneath,  will 
cause  a  reduction  of  its  inward  curvature.  These  movements 
of  the  cover  are  transmitted  and  multiplied  by  a  combination 
of  delicate  levers  that  act  on  the  index  hand  shown  in  the 
figure  and  cause  it  to  move  to  the  right  or  left  over  a  grad- 
uated scale. 

These  barometers  are  compensated  (self-correcting)  for 
variations  in  temperature.  The  better  grade  of  these  instru- 
ments have  two  graduations;  the  inner  graduation  corre- 
sponds to  the  graduations  of  the  mercurial  barometer;  that 
is,  it  reads  to  inches  of  mercury.  The  outer  graduation 
represents  elevations  above  and  below  sea  level,  the  distance 
between  each  graduation  line  representing  a  difference  in 
elevation  of  10  feet. 

These  instruments  are  made  in  various  sizes,  from  the 
size  of  a  watch  up  to  8  or  10  inches  in  diameter.  They  are 
very  portable,  occupying  but  a  small  space,  are  very  light, 
and  are  quite  delicate. 

52.  Both    the    mercurial   and   the   aneroid   barometers 
operate   on   the  same  general   principle;  viz.,    that   if   two 
opposite  forces  act  on  a  body,  it  will  be  in  equilibrium  when 
the  forces  are  equal  and  will  be  set  in  motion  when  they 
are  unequal,  provided  the  difference  in  the  magnitude  of  the 
two  forces  is  sufficient  to  overcome  the  resistance. 

The  two  styles  of  barometer  differ  from  each  other  only 
in  the  method  by  which  equilibrium  is  established.  In  the 
mercurial  barometer,  the  weight  of  the  mercurial  column 
inside  the  tube  equalizes  the  outside  air  pressure;  in  the 
aneroid  barometer,  the  outside  pressure  is  equalized  by  the 
resistance  of  the  flexible  cover  and  spring  beneath  it. 

53.  Variations  of  Pressure  at  Different  Elevations. 

With  air,  as  with  water,  the  lower  we  get  the  greater  is  the 
pressure,  and  the  higher  we  get  the  less  is  the  pressure.  At 


§  6  MECHANICS  OF  FLUIDS.  35 

the  level  of  the  sea,  the  height  of  the  mercurial  column  is 
about  30  inches;  at  5,000  feet  above  the  sea,  it  is  24.7  inches; 
at  10,000  feet  above  the  sea,  it  is  20.5  inches;  at  15,000  feet, 
it  is  16.9  inches;  at  3  miles,  it  is  16.4  inches;  and  at  6  miles 
above  the  sea  level,  it  is  8.9  inches. 

Air  being  an  elastic  fluid,  the  density  or  weight  of  the 
atmosphere  also  varies  with  the  altitude.;  that  is,  a  cubic 
foot  of  air  at  an  elevation  of  5,000  feet  above  the  sea  level 
will  not  weigh  as  much  as  a  cubic  foot  at  sea  level.  This  is 
proved  conclusively  by  the  fact  that  at  a  height  of  3£  miles 
the  mercurial  column  measures  but  15  inches,  indicating 
that  half  the  weight  of  the  entire  atmosphere  is  below  that 
height.  It  is  known  that  the  height  of  the  earth's  atmos- 
phere is  at  least  50  miles;  hence,  the  air  just  before  reaching 
the  limit  must  be  in  an  exceedingly  rarefied  state.  It  is  by 
means  of  barometers  that  great  heights  are  measured.  The 
aneroid  barometer  has  the  heights  marked  on  the  dial,  so 
that  they  can  be  read  directly.  With  the  mercurial  barom- 
eter, the  heights  must  be  calculated  from  the  reading. 

54.  Pressure  In  Different  Directions. — The  atmos- 
pheric pressure  is  everywhere  present  and  presses  all  objects 
in  all  directions  with  equal  force.  If  a  book  is  laid  upon  the 
table,  the  air  presses  upon  it  in  every  direction  with  an 
equal  average  force  of  14.7  pounds  per  square  inch.  It 
would  seem  as  though  it  would  take  considerable  force  to 
raise  a  book  from  the  table,  since,  if  the  size  of  the  book 
were  8  inches  by  5  inches,  the  pressure  upon  it  would  be 
8  X  5  X  14.7  =  588  pounds;  but  there  is  an  equal  pressure 
beneath  the  book  that  counteracts  the  pressure  on  the  top. 
It  would  now  seem  as  though  it  would  require  a  great  force 
to  open  the  book,  since  there  are  two  pressures  of  588  pounds 
each  acting  in  opposite  directions  and  tending  to  crush  the 
book ;  so  it  would,  but  for  the  fact  that  there  is  a  layer  of 
air  beneath  each  leaf,  which  acts  upwards  and  downwards 
with  a  pressure  of  14. 7  pounds  per  square  inch. 

If  a  piece  of  flat  glass  be  laid  upon  a  flat  surface  that 
has  been  previously  moistened  with  water,  it  will  require 


36  MECHANICS  OF  FLUIDS.  §  6 

considerable  force  to  separate  them;  this  is  because  the 
water  helps  to  fill  the  pores  in  the  flat  surface  and  glass 
and  thus  creates  a  partial  vacuum  between  the  glass  and 
the  surface,  thereby  reducing  the  counter  pressure  beneath 
the  glass. 

55.  Tension  of  Gases. — In  Fig.  15,  the  space  above  the 
column  of  mercury  was  said  to  be  a  vacuum,  and  it  was  also 
stated  that  if  any  gas  or  air  were  present,  it  would  expand 
and  its  tension  would  force  the  column  of  mercury  down- 
wards. If  sufficient  gas  be  admitted  to  cause  the  mercury 
to  stand  at  15  inches,  the  tension  of  the  gas  is  evidently 

14  7 

—  =7.35  pounds  per    square  inch,  since  the  pressure  of 

2 

the  outside  air,  14.7  pounds  per  square  inch,  now  balances 
only  15  instead  of  30  inches  of  mercury ;  that  is,  it  bal- 
ances only  one-half  as  much  as  it  would  if  there  were  no  gas 
in  the  tube ;  therefore,  the  pressure  (tension)  of  the  gas  in 
the  tube  is  7.35  pounds.  If  more  gas  be  forced  into  the  tube 
until  the  top  of  the  mercurial  column  is  just  level  with  the 
mercury  in  the  cup,  the  gas  in  the  tube  will  then  have  a 
tension  equal  to  the  outside  pressure  of  the  atmosphere. 
Suppose  that  the  bottom  of  the  tube  is  fitted  with  a  piston, 
and  that  the  total  length  of  the  inside  of  the  tube  is 
36  inches.  If  the  piston  be  shoved  upwards  so  that  the 
space  occupied  by  the  gas  is  18  inches  long  instead  of 
36  inches,  the  temperature  remaining  the  same  as  before, 
it  will  be  found  that  the  tension  of  the  gas  within  the  tube 
is  29.4  pounds.  It  will  be  noticed  that  the  volume  occupied 
by  the  gas  is  only  half  that  in  the  tube  before  the  piston 
was  moved,  while  the  pressure  is  twice  as  great,  since 
14.7  X  2  =  29.4  pounds.  If  the  piston  be  shoved  up,  so  that 
the  space  occupied  by  the  gas  is  only  9  inches  instead  of 
18  inches,  the  temperature  still  remaining  the  same,  the 
pressure  will  be  found  to  be  58.8  pounds  per  square  inch. 
The  volume  has  again  been  reduced  one-half  and  the  pres- 
sure increased  twofold,  since  29.4  X  2  —  58.8  pounds.  The 
space  now  occupied  by  the  gas  is  9  inches  long,  whereas, 


§  6  MECHANICS  OF  FLUIDS.  37 

before  the  piston  was  moved,  it  was  36  inches  long;  as  the 
tube  is  assumed  to  be  of  uniform  diameter  throughout  its 
length,  the  volume  is  now  -^  =  ^  of  its  original  volume,  and 

its  pressure  is  — —  =  4  times  its  original  pressure.  More- 
over, if  the  temperature  of  the  confined  gas  remains  the 
same,  the  pressure  and  volume  will  always  vary  in  a  similar 
way. 

56.  Absolute  and  Gauge  Pressures. — From  the  above 
explanation,  it  will  be  apparent  that  the  pressure  in  the  tube 
is  the  pressure  above  that  of  a  perfect  vacuum.     Pressures 
reckoned    above    vacuum   are   called   absolute    pressures. 
The  only  pressure  gauges  that  indicate  absolute  pressures 
are  the  mercurial  and  aneroid  barometers;  ordinary  pressure 
gauges,  such  as  the  common  steam  gauge,  indicate  pressures 
above  the  pressure  of  the  atmosphere.     Pressures  above  that 
of  the  atmosphere  are  commonly  called  gauge  pressures. 
Gauge  pressures  are  changed  to  absolute  pressures  by  adding 
14.7  pounds  per  square  inch  to  their  readings.     Truly  speak- 
ing,   the   pressure   indicated   by  a    barometer,  reduced   to 
pounds   pressure,  should  be  added  to  the  gauge  pressure, 
since  the  value  14. 7  pounds  only  represents  the  mean  atmos- 
pheric pressure  under  normal  conditions  at  sea  level. 

57.  Mariotte's  Law. — The  law  that  states  the  effect  of 
compressing  and  expanding  gases  is  called  Mariotte's  law 

and  is  as  follows : 

The  temperature  remaining  the  same,  the  volume  of  a  given 
quantity  of  gas  varies  inversely  as  the  absolute  pressure. 

The  meaning  of  the  law  is  this:  If  the  volume  of  a  gas 
be  diminished  to  £,  £,  \,  etc.  of  its  former  value,  the  ten- 
sion will  be  increased  2,  3,  5,  etc.  times,  or  if  the  outside 
pressure  be  increased  2,  3,  5,  etc.  times,  the  volume  of  the 
gas  will  be  diminished  to  £,  £,  ^,  etc.  of  its  original  volume, 
the  temperature  remaining  constant.  It  also  means  that  if 
a  gas  is  under  a  certain  pressure,  and  this  pressure  is  dimin- 
ished to  £,  £,  -j1^,  etc.  of  its  original  intensity,  the  volume  of 


38  MECHANICS  OF  FLUIDS.  §  6 

the  confined  gas  will  be  increased  2,   4,  10,  etc.  times — its 
tension  decreasing  at  the  same  rate. 

Suppose  3  cubic  feet  of  air  to  be  under  a  pressure  of 
60  pounds  per  square  inch  in  a  cylinder  fitted  with  a  mov- 
able piston ;  then  the  product  of  the  volume  and  pressure  is 
3  x  60  =  180.  Let  the  volume  be  increased  to  6  cubic  feet; 
then'  the  pressure  will  be  30  pounds  per  square  inch,  and 
30  x  6  =  180,  as  before.  Let  the  volume  be  increased  to 
24  cubic  feet ;  it  is  then  -2^4-  =  8  times  its  original  volume, 
and  the  pressure  is  £  of  its  original  pressure,  or  60  X  £ 
=  7£  pounds,  and  24  X  7£  =  180,  as  in  the  two  preceding 
cases.  It  will  now  be  noticed  that  if  a  gas  be  enclosed  within 
a  confined  space  and  allowed  to  expand  without  losing  any 
heat,  the  product  of  the  pressure  and  the  corresponding  volume 
for  any  one  position  of  the  piston  is  the  same  as  for  any  other 
position.  If  the  piston  were  forced  inwards  so  as  to  com- 
press the  air,  the  same  results  would  be  obtained. 

58.  Application  of  Mariotte's  Law. — If  the  volume 
of  the  vessel  and  the  pressure  of  the  gas  are  known,  and  it  is 
desired  to  know  the  pressure  after  the  first  volume  has  been 
changed : 

Rule  7. — Divide  the  product  of  the  first,  or  original, 
volume  and  pressure  by  the  new  volume;  the  result  will  be  the 
new  pressure. 

Or,  let     /   =  original  absolute  pressure ; 
/,  =  final  absolute  pressure; 
v  =  volume  corresponding  to  the  pressure  / ; 
vl  =  volume  corresponding  to  the  pressure  /r 

Then,  /,=£?. 

», 

EXAMPLE. — At  the  point  of  cut-off  in  a  steam  engine,  the  amount  of 
steam  in  the  cylinder  is  862  cubic  inches.  The  pressure  at  this  point  is 
120  pounds  per  square  inch.  What  will  be  the  pressure  of  the  steam 
when  the  piston  has  reached  the  end  of  its  stroke  and  the  volume  is 
1,800  cubic  inches  ? 


§  6  MECHANICS  OF  FLUIDS.  39 

SOLUTION.  —  Applying  the  rule,  we  get 

'    A  =  86i  ^r!20  =  57-47  lb-  Per  sq-  in-  absolute.     Ans. 

l,oUU 

59.  If  it  is  required  to  determine  the  volume  after  a 
change  in  the  pressure: 

Rule  8.  —  Divide  the  product  of  tJie  original  volume  and 
pressure  by  the  new  pressure;  the  result  will  be  the  new 
volume. 

Or,  using  the  same  letters  as  before, 

- 


EXAMPLE  1.—  At  the  commencement  of  compression,  the  volume  of 
the  steam  is  380  cubic  inches  and  the  pressure  is  18  pounds  per  square 
inch.  At  the  end  of  compression,  the  pressure  is  112  pounds  per  square 
inch.  What  is  the  final  volume  ? 

SOLUTION.—  Applying  the  rule,  we  get 
380  X  18 


112 


=  61.07  cu.  in.     Ans. 


EXAMPLE  2. — A  vessel  contains  10  cubic  feet  of  air  at  a  pressure  of 
15  pounds  per  square  inch  and  has  25  cubic  feet  of  air  of  the  same 
pressure  forced  into  it ;  what  is  the  resulting  pressure  ? 

SOLUTION. — The  original  volume  =  10  +  25  =  35  cubic  feet  ;  the 
original  pressure  is  15  pounds  per  square  inch;  the  final  volume  is 
10  cubic  feet.  Hence,  applying  rule  7,  we  get 

OK     \/    1  5 

p\  =  — ^r —  =  52.5  lb.  per  sq.  in.  absolute.     Ans. 

It  must  be  remembered  that  in  the  preceding  examples 
the  temperature  is  supposed  to  remain  constant.  When  the 
temperature  changes  during  expansion  and  compression,  the 
problem  becomes  a  rather  difficult  one  and  cannot  be  solved 
by  ordinary  arithmetic. 


EXAMPLES  FOR  PRACTICE. 

1.  A  vessel  contains  25  cubic  feet  of  gas  at  a  pressure  of  18  pounds 
per  square  inch ;  if  125  cubic  feet  of  gas  having  the  same  pressure  are 
forced  into  the  vessel,  what  will  be  the  resulting  pressure  ? 

Ans.  108  lb.  per  sq.  in. 

H.  8.    L— 20 


40  MECHANICS  OF  FLUIDS.  §  6 

2.  The  volume  of  steam  in  the  cylinder  of  a  steam  engine  at  cut-off 
is  1.85  cubic  feet  and  the  pressure  is  85  pounds  per  square  inch;  the 
pressure  at  the  end  of  the  stroke  is  25  pounds  per  square  inch.     What 
is  the  new  volume?  Ans.  4.59  cu.  ft. 

3.  A  receiver  contains    180  cubic  feet  of   gas  at  a  pressure  of 
20  pounds  per  square  inch ;  if  a  vessel  holding  12  cubic  feet  be  filled 
from  the  larger  vessel  until  its  pressure  is  20  pounds  per  square  inch, 
what  will  be  the  pressure  in  the  larger  vessel  ? 

Ans.  18£  Ib.  per  sq.  in. 

4.  A  spherical  shell  has  a  part  of  the  air  within  it  removed,  forming 
a  partial  vacuum ;  if  the  outside  diameter  of  the  shell  is  18  inches  and 
the  pressure  of  the  air  within  is  5  pounds  per  square  inch,  what  is  the 
total  pressure  tending  to  crush  the  shell  ?  Ans.  9,873.42  Ib. 


PNEU1VIATIC    MACHINES. 


THE  AIR  PUMP. 

6O.     The  air  pump  is  an  instrument  for  removing  air 
from  a  given   space.     A  section   of   the   principal  parts  is 


FIG.  19. 


shown  in  Fig.  19  and  the  complete  instrument  in  Fig.  20. 
The  closed  vessel  R  is  called  the  receiver,  and  the  space 


MECHANICS  OF  FLUIDS. 


41 


that  it  encloses  is  that  from  which  it  is  desired  to  remove 
the  air.  It  is  usually  made  of  glass,  and  the  edges  are 
ground  so  as  to  be  perfectly  air-tight.  When  made  in 
the  form  shown,  it  is  called  a  bell-jar  receiver.  The 
receiver  rests  upon 
a  horizontal  plate,  in 
the  center  of  which  is 
an  opening  communi- 
cating with  the  pump 
cylinder  C  by  means  of 
the  passage  tt.  The 
pump  piston  accu- 
rately fits  the  cylinder, 
and  has  a  valve  V 
opening  upwards. 
Where  the  passage  // 
joins  the  cylinder, 
another  valve  V  is 
placed,  which  also 
opens  upwards.  When 
the  piston  is  raised,  the 
valve  V  closes,  and, 
since  no  air  can  get 
into  the  cylinder  from 

above,  the  piston  leaves  a  vacuum  behind  it.  The  pressure 
upon  V  being  now  removed,  the  tension  of  the  air  in  the 
receiver  R  causes  V  to  rise;  the  air  in  the  receiver  and 
passage  1 1  then  expands  so  as  to  occupy  the  additional  space 
provided  by  the  upward  movement  of  the  piston. 

The  piston  is  now  pushed  down,  the  valve  V  closes,  the 
valve  V  opens,  and  the  air  in  C  escapes.  The  lower 
valve  V  is  sometimes  supported,  as  shown  in  Fig.  19,  by  a 
metal  rod  passing  through  the  piston  and  fitting  it  some- 
what tightly.  When  the  piston  is  raised  or  lowered,  this 
rod  moves  with  it..  A  button  near  the  upper  end  of  the 
rod  confines  its  motion  within  very  narrow  limits,  the  pis- 
ton sliding  upon  the  rod  during  the  greater  part  of  the 
journey.  In  the  complete  form  of  the  instrument  shown 


42  MECHANICS  OF  FLUIDS.  §  6 

in  Fig.   20,  communication  between  receiver  and  pump  is 
made  by  means  of  the  tube  /. 

61.  Degrees  and  limits  of  Exhaustion. — Suppose  that 
the  volume  of  R  and  t  together  is  four  times  that  of  C, 
Fig.  19,  and  that  there  are,  say,  200  grains  of  air  in  R  and  /, 
and  50  grains  in  C  when  the  piston  is  at  the  top  of  the  cylin- 
der. At  the  end  of  the  first  stroke,  when  the  piston  is  again 
at  the  top,  50  grains  of  air  in  the  cylinder  C  will  have  been 
removed,  and  the  200  grains  in  R  and  t  will  occupy  the 
space  R,  t,  and  C.  The  ratio  between  the  sum  of  the  spaces 
R  and  t  and  the  total  space  R  -f  t  +  C  is  £ ;  hence,  200 
X  -|  =  160  grains  =  the  weight  of  air  in  R  and  t  after  the  first 
stroke.  After  the  second  stroke,  the  weight  of  the  air  in  R 
and  t  would  be  (200  X  |)  X  f  =  200  X  (|)s  =  200  X  if  =  128 
grains.  At  the  end  of  the  third  stroke,  the  weight  would  be 
[200  X  (|)2]  X  -£  =  200  X  (f)s  =  200  X  T6/T  =  102.4  grains. 
At  the  end  of  n  strokes  the  weight  would  be  200  X  (£)".  It 
is  evident  that  it  is  impossible  to  remove  all  the  air  that  is 
contained  in  R  and  t  by  this  method.  It  requires  an  exceed- 
ingly good  air  pump  to  reduce  the  tension  of  the  air  in  R  to 
-5*5-  inch  of  mercury.  When  the  air  has  become  so  rarefied 
as  this,  the  valve  V  will  not  lift,  and,  consequently,  no  more 
air  can  be  exhausted. 

63.  The  Dashpot. — The  pressure  of  the  atmosphere  is 
utilized  in  Corliss  engines  to  close  the  steam  valves  rapidly. 
For  this  purpose  a  so-called  dashpot  is  used.  The  dash- 
pot  of  a  Bullock-Corliss  engine  is  shown  in  two  positions  in 
Fig.  21.  It  consists  of  the  base  A,  which  is  fastened  to  the 
floor,  and  a  plunger  B  pivoted  to  the  end  of  a  crank-arm 
keyed  to  the  stem  of  the  steam  valve.  The  base  is  accurately 
turned  and  bored  to  fit  the  plunger;  packing  rings  are 
fitted  to  both  the  plunger  and  the  base  in  order  to  make  an 
air-tight  joint. 

The  annular  space  around  the  central  stationary  piston  is 
in  communication  with  the  outside  air  by  a  small  passage, 
the  opening  of  which  can  be  increased  or  decreased,  at  will, 


MECHANICS  OF  FLUIDS. 


43 


by  means  of  the  valve  C.     In  operation,  the  plunger  is  first 
in  the  position  shown  in  Fig.   21  (b}.     It  is  then  picked  up 

by  the  valve  gear  of 
the  engine  and  drawn 
up  until  it  is  in  the 
position  shown  in 
Fig.  21  (a).  The  large 
piston  on  the  lower 
end  of  the  plunger 
is  in  communication 
with  the  atmosphere 
through  the  valve  C, 
and  hence  the  pres- 
sure of  the  atmos- 
phere on  both  sides 
of  the  piston  is  equal. 
The  upper  part  of 
the  plunger,  however, 
has  an  equal  pressure 
on  both  its  sides  only 
when  it  is  down,  as 
in  Fig.  21  (b).  When 
it  is  drawn  up,  the 
air  in  the  space  D 
expands  and  a  partial 
vacuum  is  formed. 
The  valve  gear  next 
releases  the  plunger, 
and  as  the  atmos- 
pheric pressure  is 
much  greater  than 
the  pressure  within  D, 
the  plunger  rapidly 
descends.  During  its 
descent,  the  large  pis- 
ton at  the  end  of 
the  plunger  gradually 
compresses  the  air  in  the  annular  space  beneath  it,  and  is 


44 


MECHANICS  OF  FLUIDS. 


thus  gradually  brought  to  rest.  The  valve  C  serves  to 
regulate  the  amount  of  compression  and  at  the  same  time 
admits  air  during  the  up  stroke  of  the  plunger. 


THE   SIPIIOX. 

63.     Theory  of  the  Siphon. — The  action  of  the  siphon 

illustrates  the  effect  of  atmospheric  pressure.     It  is  simply 
^s^L.^  a    bent    tube    having    unequal 

Jf^  ^"^^V  branches,    open    at    both    ends, 

if  \  and  is  used  to  convey  a  liquid 

from  a  higher  point  to  a  lower 
over  an  intermediate  point  that 
is  higher  than  either.  In  Fig.  22, 
A  and  B  are  two  vessels,  B 
being  lower  than  A,  and  A  C  B 
is  the  bent  tube,  or  siphon. 
Suppose  this  tube  to  be  filled 
with  water  and  placed  in  the 
vessels  as  shown,  with  the  short 
branch  A  C  in  the  vessel  A. 
The  water  will  flow  from  the 
vessel  A  into  the  vessel  B  as 
long  as  the  level  of  the  water 
FIG- 22-  in  B  is  below  the  level  of  the 

water  in  A  and  the  level  of  the  water  in  A  is  above  the  lower 
end  of  the  tube  A  C.  The  atmospheric  pressure  on  the  sur- 
faces of  A  and  B  tends  to  force  the  water  up  the  tubes  A  C 
and  B  C.  When  the  siphon  is  filled  with  water,  each  of  these 
pressures  is  counteracted  in  part  by  the  pressure  of  the  water 
in  that  branch  of  the  siphon  that  is  immersed  in  the  water 
on  which  the  pressure  is  exerted.  The  atmospheric  pressure 
opposed  to  the  weight  of  the  longer  column  of  water  will, 
therefore,  be  resisted  more  strongly  than  that  opposed  to  the 
weight  of  the  shorter  column ;  consequently,  the  pressure 
exerted  on  the  shorter  column  will  be  greater  than  that  on 
the  longer  column,  and  this  excess  of  pressure  will  produce 
motion. 


§  G  MECHANICS  OF  FLUIDS.  45 

64.  Let  A  —  the  area  of  the  tube ; 

h  =  D  C  =  the  vertical  distance  between  the 
surface  of  the  water  in  B  and  the  highest 
point  of  the  center  line  of  the  tube ; 

kl  =  E  C  —  the  distance  between  the  surface 
of  the  water  in  A  and  the  highest  point  of 
the  center  line  of  the  tube. 

The  weight  of  the  water  in  the  short  column  is  .03617 
X  A  /;,,  and  the  resultant  atmospheric  pressure  tending  to 
force  the  water  up  the  short  column  is  14.7  X  A  —  .03617 
X  A  h^.  The  weight  of  the  water  in  the  long  column  is 
.03617  A  /z,  and  the  resultant  atmospheric  pressure  tending 
to  force  the  water  up  the  long  column  is  14. 7  A  —  .03617  A  h. 
The  difference  between  these  two  is  (14.7  A  —  ,03617  A  k^ 
-  (14.7  A  —  .03617  A  h)  —  .03617  A  (k  —  //,).  But//  —  //, 
=  E  D  =  the  difference  between  the  levels  of  the  water  in 
the  two  vessels.  In  the  above,  h  and  //1  were  taken  in  inches 
and  A  in  square  inches. 

It  will  be  noticed  that  the  short  column  must  not  be  higher 
than  34  feet  for  water,  or  the  siphon  will  not  work,  since  the 
pressure  of  the  atmosphere  will  not  support  a  column  of  water 
that  is  higher  than  34  feet. 

65.  Fig.  23  shows  the  actual  construction  of  a  siphon. 
It  is  desired  to  convey  the  water  from  D  to  E.     The  point 
of   discharge  must  always   be  lower  than  the  point  from 
which  the  water  is  taken,  otherwise  a  siphon  cannot  work. 
The  siphon  consists  of  ordinary  iron  pipe  jointed  in  any 
convenient   manner  so  as   to   be   air-tight.       It   has   three 
valves  A,  B,  and  C.     The  suction  end  is  provided  with  a 
perforated  strainer  c  to  keep  out  large  particles  of  foreign 
matter. 

In  order  to  start  the  siphon,  it  is  necessary  to  remove  the 
air  in  the  pipe.  This  is  done  by  closing  the  valves  A  and  B 
and  opening  the  valve  C,  which  should  always  be  located  at 
the  highest  point  of  the  siphon.  Water  is  then  poured  into 
the  funnel  above  the  valve  C  until  the  pipe  is  filled.  When 


46  MECHANICS  OF  FLUIDS.  §  6 

no  more  water  can  be  poured  in,  that  is,  when  all  the  air 
has  been  driven  out,  the  valve  C  is  closed  and  the  valves  A 
and  B  are  opened ;  the  siphon  will  now  start. 


FIG.  23. 

In  practice,  it  has  been  found  that  the  distance  s  must 
not  exceed  28  feet  at  sea  level,  and  that  the  siphon  will  work 
more  satisfactory  if  it  is  less.  The  greater  the  distance  h 
between  the  two  water  levels,  the  better  the  siphon  will  work. 

66.  In  order  that  a  siphon  will  work  properly,  it  is  nec- 
essary that  air  should  be  kept  out  of  the  pipe,  or,  if  it  does 
get  in,  means  should  be  provided  for  its  escape.     The  joints 
of  the  pipe  must  be  perfectly  air-tight.     But  some  air  will 
always  be  carried  in  with  the  water,  and  this  air  will  collect 
at  the  highest  point.     The  bad  effects  of  this  can  be  mini- 
mized by  having  a  straight  horizontal  pipe  at  the  highest 
point  instead  of  a  sharp  bend. 

67.  A    device   that  will    remove  the   air  is    shown    in 
Fig.  24.     Here  A  is  an  air-tight  vessel  connected  with  the 
siphon  by  two  small  pipes  B  and  C.   The  pipe  B  extends  near- 
ly to  the  top  of  A,  while  the  pipe  C  barely  enters  the  bot- 
tom.   Each  pipe  has  a  valve  D  and  E.    A  valve  F  and  funnel  G 
are  placed  on  top  of  the  vessel.     When  the  siphon  ceases  to 
flow,  which  is  an  indication  that  air  has  collected  (it  is  here 


MECHANICS  OF  FLUIDS. 


47 


assumed  that  the  suction  end  has  not  become  uncovered), 
the  valves  D  and  E  are  closed  and  the  valve  F  is  opened. 
The  vessel  is  now  filled  with  water,  the  valve  F  is  closed, 
and  D  and  E  are  opened.  The  water  will  now  flow  through 
C  into  the  siphon  and  force  out  the  air  collected  there, 
which  passes  through  B  to  the  top  of  A.  When  all  air  is 
out  of  the  siphon,  D  and  E  are  shut  and  F  is  opened.  The 
vessel  A  is  now  filled  with  water,  F  is  shut,  D  and  E  are 


opened  and  left  open.  Any  air  that  enters  the  siphon  will, 
instead  of  collecting  at  //,  ascend  the  pipe  B  and  force  out 
a  small  amount  of  water  through  C.  This  will  continue 
until  A  is  filled  with  air,  when  the  valves  D  and  E  should  be 
closed  and  A  refilled.  This  arrangement  may  also  be  used 
to  fill  the  siphon  for  the  purpose  of  setting  it  to  work. 

The  highest  point  of  the  water  in  A  should  not  be  more 
than  28  feet  at  sea  level  above  the  water  level  at  the  suction 
end. 


PUMPS. 

68.  A  pump  is  a  machine  for  conveying  liquids  from 
one  level  to  a  higher  level  or  for  performing  work  equiva- 
lent to  such  an  operation. 


48 


MECHANICS  OF  FLUIDS. 


69.  Classification  of  Pumps. — Pumps  may  be  divided 
into  three  general  divisions,  according  to  the   service  they 
perform,  viz.,  suction  pumps,  lifting  pumps,  and  force  pumps. 
They  may  also  be  divided  into  two  general  classes,  single- 
acting  and    double-acting  pumps,    according   as    they    take 
water  on  one  side  or  on  both   sides   of  the  water   piston. 
According  to  the  arrangement  of  the  pump  cylinders,  they 
are   classified    as   simple,    duplex,    or    triplex   pumps.       As 
pumps  displace  the  liquids  in  various  ways,  they  may  also 
be  divided  according  to  the  method  of  displacement,  into 
reciprocating,  centrifugal,  and  rotary  pumps.     Reciprocating 
pumps  only  will  be  considered  here. 

70.  The   Suction   Pump. — A   section    of   an   ordinary 
suction  pump  is  shown  in  diagrammatic  form  in  Fig.  25. 

Suppose  the  piston, 
or  bucket  as  it  is 
commonly  terned,  to 
be  at  the  bottom  of 
the  cylinder  and  to 
be  just  on  the  point 
of  moving  upwards 
in  the  direction  of  the 
arrow.  As  the  piston 
rises  it  leaves  a  par- 
tial vacuum  behind  it, 
and  the  atmospheric 
pressure  on  the  sur- 
face of  the  water  in 
the  well  causes  it  to 
rise  in  the  pipe  P  for 


the  same  reason  that 
FIG-  25-  the   mercury  rises  in 

the  barometer  tube.  The  water  rushes  up  the  pipe  and 
lifts  the  suction  valve  V,  filling  the  empty  space  in  the  cylin- 
der B  caused  by  the  displacement  of  the  piston.  When  the 
piston  has  reached  the  end  of  its  stroke,  the  water  entirely 
fills  the  space  between  the  bottom  of  the  piston  and  the 


§  6  MECHANICS  OF  FLUIDS.  49 

bottom  of  the  cylinder,  and  also  the  pipe  P.  The  instant  the 
piston  begins  its  down  stroke,  the  water  in  the  chamber  B 
begins  to  flow  back  into  the  well,  and  its  downward  flow 
forces  the  valve  V  to  its  seat,  thus  preventing  any  further 
escape  of  the  water.  As  the  piston  descends,  the  water 
must  give  way  to  it,  and  since  the  suction  valve  V  is  closed, 
the  bucket  valves  u,  u  must  open,  and  thus  allow  the  water 
to  pass  through  the  piston,  as  shown  in  the  right-hand 
figure.  When  the  piston  has  reached  the  end  of  its  down- 
ward stroke  and  commenced  its  upward  movement  again, 
the  water  flowing  through  the  piston  quickly  closes  the 
valves  tt,  u.  All  the  water  resting  on  the  top  of  the  piston 
is  then  lifted  by  the  piston  on  its  upward  stroke  and  dis- 
charged through  the  spout  A  ;  the  valve  V  again  opens  and 
the  water  fills  the  space  below  the  piston,  as  before. 

71.  It  is  evident  that  the  distance  between  the  piston 
when  at  the  top  of  its  stroke  and  the  surface  of  the  water  in 
the  well  must  not  exceed  34  feet,  the  highest  column  of 
water  that  the  pressure  of  the  atmosphere  will  sustain,  since, 
otherwise,  the  water  in  the  pipe  would   not  rise  and  fill  the 
cylinder  as  the  piston  ascended.     In  practice,  this  distance 
should  not  exceed  28  feet.     This  is  due  to  the  fact  that  there 
is  a  little  air  left  between  the  bottom  of  the  piston  and  the 
bottom  of  the  cylinder,  a  little  air  leaks  through  the  valves, 
which  are  not  perfectly  air-tight,  and  a  pressure  is  needed 
to  raise  the  valve  against  its  own  weight,  which,  of  course, 
acts  downwards. 

There  are  many  varieties  of  the  suction  pump,  differing 
principally  in  the  construction  of  the  valves  and  piston,  but 
the  principle  is  the  same  in  all. 

72.  The  Lifting  Pump. — In  some  cases  it  is  desired  to 
raise  water  higher  than  it  can  be  forced  by  the  pressure  of 
the  atmosphere  into  the  chamber  of  a  simple  suction  pump, 
such  as  is  shown  in  Fig.  25.      To  accomplish  this  the  pump 
chamber  with  its  bucket  and  valves  are  set  at  a  distance 
above  the  supply  not  exceeding  that  to  which  the  air  will 
successfully  force  the  water.     A  closed  pipe  P\  Fig.  26, 


50 


MECHANICS  OF  FLUIDS. 


called  the  delivery,  or  discharge,  pipe,  is  then  led  from  the 
upper  part  of  the  chamber  to  the  point  where  it  is  desired 
to  deliver  the  water.  Such  a  pump  is  often  called  a  lifting 
pump. 

In  order  to  prevent  the  leakage  of  water  around  the  pis- 
ton rod,  a  stuffmgbox  5  is  provided.  The  lower  end  of  the 
discharge  pipe  P'  is  sometimes  fitted  with  a  valve  c  to  pre- 
vent the  water  flowing  back  into  the  pump  chamber;  this 


valve  is  not  essential  to  the  operation  of  the  pump,  how- 
ever, since  the  valve  V  prevents  the  water  in  the  chamber 
and  discharge  pipe  discharging  during  the  downward  motion 
of  the  piston. 

While  it  is  customary  to  consider  lifting  pumps  and  suc- 
tion pumps  as  two  different  types  of  pumps,  there  is  in 
reality  no  difference  in  their  operation,  as  a  careful  study  of 
Figs.  25  and  26  will  show.  The  only  difference  is  that  the 
water  is  discharged  by  a  suction  pump  at  the  level  of  the 


MECHANICS  OF  FLUIDS. 


51 


pump,  while  a  lifting  pump  discharges  the  water  above  the 
level  of  the  pump. 

73.  Force  Pumps. — The  force  pump  differs  from  the 
lifting  pump  in  one  important  particular,  that  is,  in  the  fact 
that  its  piston  is  solid.     A  section  of  a  force  pump  is  shown 
in  Fig.   27.     As  the  piston  ascends,  as  shown  in  the  left- 
hand  figure,  the  pressure  of  the  atmosphere  forces  the  water 
up  the  suction  pipe  P\  the  water  opens  the  suction  valve  V 
and  flows  into  the  pump  cylinder.     When  the  piston  moves 
down,  as  shown  in  the  right-hand  figure,  the  suction  valve 
is  closed  and  the  delivery  valve  V  opened.     The  water  in 
the  pump  cylinder  is  now  forced  up  the  delivery  pipe  P' . 
When  the  piston  again  begins  to  move  upwards,  the  deliv- 
ery valve  is  closed  by  the  water  above  it  and  the  suction 
valve  opened  by  the  pressure  of  the  atmosphere  on  the  water 
below  it. 

74.  Plunger  Pumps. — When  force  pumps  are  used  to 
convey  water  to  great   heights  or  to  force  water  against 
heavy  pressures,  the 

great  pressure  of 
the  water  makes  it 
extremely  difficult 
to  keep  the  water 
from  leaking  past 
the  piston,  and  the 
constant  repairing 
and  renewal  of  the 
piston  packing  be- 
comes a  nuisance 
and  involves  serious 
expense.  To  obvi- 
ate this  drawback, 
plunger  pumps  have 
been  designed,  one 
of  which  is  shown 
diagrammatically  in 
Fig.  28.  The  action 


FIG.  28. 


MECHANICS  OF  FLUIDS. 


does  not  differ  in  any  way  from  that  of  the  piston  force 
pump.  During  the  up  stroke  of  the  plunger,  the  suction 
valve  is  open  and  the  delivery  valve  is  closed;  during  the 
down  stroke,  the  suction  valve  is  closed  and  the  delivery 
valve  is  open. 

75.  The  force  pumps  shown  so  far  are  single-acting, 
that  is,  the  water  is  forced  into  the  delivery  pipe  only  during 
the  downstroke  or  forward  stroke  of  the  piston  or  plunger. 
Force  pumps,  either  of  the  piston  or  plunger  pattern,  may 
be  constructed  so  as  to  force  water  into  the  delivery  pipe 
both  during  the  forward  and  return  stroke.      They  are  then 
called  double-acting. 

76.  A  double-acting  force  pump  of  the  piston  pattern 
is  shown  in  Fig.  29.     Such  a  pump  has  two  sets  of  suction 

valves  and  delivery 
valves,  one  set  for  each 
side  of  the  piston. 
With  the  piston  mov- 
ing in  the  direction  of 
the  arrow,  the  pressure 
of  the  atmosphere 
forces  the  water  up  the 
suction  pipe  P  into 
the  left-hand  end  of 
the  pump  cylinder,  the 
left-hand  suction  valve 
opens  and  the  left-hand 
delivery  valve  is  closed. 
P  The  piston  in  moving 

FlG-  ™-  the  water  in  the  right- 

hand  end  of  the  pump  cylinder;  as  a  consequence  the 
right-hand  suction  valve  is  closed  and  the  right-hand  deliv- 
ery valve  opens.  The  water  now  flows  up  the  delivery 
pipe  P'.  Imagine  that  the  piston  is  at  the  end  of  its  stroke 
and  commences  to  move  to  the  left.  Its  first  movement 


§  6  MECHANICS  OF  FLUIDS.  53 

promptly  closes  the  left-hand  suction  valve  and  opens  the 
left-hand  delivery  valve.  It  also  closes  the  right-hand 
delivery  valve  and  opens  the  right-hand  suction  valve. 

It  is  thus  seen  that  with  the  arrangement  given,  which 
shows  the  principle  of  operation  of  all  double-acting  pumps, 
the  piston  will  discharge  water  both  during  the  forward  and 
the  return  stroke.  While  the  pump  shown  is  a  horizontal 
pump,  it  may  be  vertical  as  well. 


STRENGTH  OF  MATERIALS. 


GENERAL  PRINCIPLES. 

1.  When  a  force  is  applied  to  a  body,  it  changes  either 
its  form  or  its  volume.     A  force,  when  considered  with  ref- 
erence to  the  internal  changes  it-  tends  to  produce  in  any 
solid,  is  called  a  stress. 

Thus,  if  we  suspend  a  weight  of  2  tons  by  a  rod,  the  stress 
in  the  rod  is  2  tons.  This  stress  is  accompanied  by  a  length- 
ening of  the  rod,  which  increases  until  the  internal  stress  or 
resistance  is  in  equilibrium  with  the  external  weight. 

2.  .Classification  of  Stresses. — Stresses  may  be  classi- 
fied   as   follows :    Tensile,    or  pulling  stress ;  transverse,  or 
bending  stress  ;  compressive,  or  pushing  stress ;  shearing,  or 
cutting  stress  ;  torsional,  or  twisting  stress. 

3.  A  unit  stress  is  the  amount  of  stress  on  a  unit  of 
area,  and  may  be  expressed   either  in  pounds  or  tons  per 
square  inch  or  per  square  foot ;  or,  it  is  the  load  per  square 
inch  or  per  square  foot  on  any  bod)'-. 

Thus,  if  10  tons  are  suspended  by  a  wrought-iron  bar  that 
has  an  area  of  5  square  inches,  the  unit  stress  is  2  tons  per 
square  inch,  because  J/  =  2  tons. 

4.  Strain  is  the  deformation  or  change  of  shape  of  a 
body  resulting  from  stress. 

For  example,  if  a  rod  100  feet  long  is  pulled  in  the  direc- 
tion of  its  length,  and  if  it  is  lengthened  1  foot,  it  is  strained 
-j-Jrj-  its  length,  or  1  per  cent. 

For  notice  of  the  copyright,  see  page  immediately  following  the  title  page. 
H.    8.  I.-21 


2  STRENGTH  OF  MATERIALS.  §  7 

5.  Elasticity  is  the  property  by  virtue  of  which  a  body 
regains  its  original  form  after  the  external   force   on   it   is 
withdrawn,  provided  the  stress  has  not  exceeded  the  elastic 
limit.     It  is  a  property  possessed  by  all  bodies. 

Consequently,  we  see  from  this  that  all  material  is  length- 
ened or  shortened  when  subjected  to  either  tensile  or  com- 
pressive  stress,  and  the  change  of  the  length  within  the 
elastic  limit  is  directly  proportional  to  the  stress. 

For  stresses  within  the  elastic  limits,  materials  are  per- 
fectly elastic,  and  return  to  their  original  length  on  removal 
of  the  stresses;  but  when  their  elastic  limits  are  exceeded, 
the  changes  of  their  lengths  are  no  longer  regular,  and  a 
permanent  set  takes  place.  The  destruction  of  the  material 
has  then  begun. 

6.  The  measure  of  elasticity  of  any  material  is  the 
change  of  length  under  stress  within  the  elastic  limit. 

7.  The  elastic  limit  is  that  unit  stress  under  which  the 
permanent  set  becomes  visible. 

8.  The  elasticity  of  wrought  iron  and  of  all  grades  of 
steel    is   practically  the    same;    that    is,    within    the   clastic 
limit  each  material  will  change   an   equal  amount  of  length 
under  the  same  stress.     The  elastic  limit,  however,  is  not 
the  same  for  steel  as  for  iron ;  it  is  higher  for  soft  steel  than 
for  wrought  iron,  and,  in  general,  the  harder  and  stronger 
the  steel  the  higher  will  be  its  elastic  limit.     As  a   conse- 
quence, steel  will   lengthen  or  shorten  more  than  wrought 
iron,  and  hard  steel  more  than  soft,  before  its  elasticity  or 
ability  to  return  to  its  original  dimensions  is  injured. 


TENSILE   STRENGTH  OF  MATERIALS. 

9.  The  tensile  strength  of  any  material  is  the  resistance 
offered  by  its  fibers  to  being  pulled  apart. 

10.  The  tensile  strength  of  any  material   is  proportional 
to  the  area  of  its  cross-section. 


STRENGTH  OF  MATERIALS. 


3 


Consequently,  when  it  is  required  to  find  the  safe  tensile 
strength  of  any  material,  we  have  only  to  find  the  area  at 
the  minimum  cross-section  of  the  body,  and  multiply  it  by 
the  load  per  square  inch  that  it  can  safely  carry,  as  given  in 
the  following  table  under  the  heading  "  Safe  Loads." 

NOTE. — The  minimum  cross-section  referred  to  in  the  above  para- 
graph is  that  section  of  the  material  which  is  pierced  with  holes,  such 
as  bolt  or  rivet  holes  in  iron,  or  knots  in  wood,  if  there  are  any. 


TABLE  I. 


TEXSIL.E  STRENGTH  OF  MATERIALS. 


Material. 

Ultimate  Tensile 
Strength.    Pounds 
per  Square  Inch. 

Safe  Loads.     Pounds 
per  Square  Inch. 

Sudden. 

Gradual. 

Steady. 

Timber 

10,000 
16,000-  20,000 
30,000-  35,000 
45,000-  55,000 
45,000-  55,000 
52,000-  62,000 
52,000-  62,000 
60,000-  75,000 
75,000-  90,000 

90,000-115,000 
100,000-125,000 
125,000-180,000 
14,000-  20,000 
30,000-  36,000 
70,000-  80,000 

30,000-  40,000 
30,000-  36,000 

600 
1,600 
2,000 
4,500 
6,000 
6,000 
6,000 
6,500 
7,000 

7,000 
8,000 
15,000 
1,400 
3,000 
7,000 

2,500 
3,500 

900 
2,400 
3,000 
9,000 
10,000 
10,000 
10,000 
11,000 
14,000 

14,000 
15,000 
23,000 
2,200 
5,000 
11,000 

3,700 
6,000 

1,200 
3,200 
4,000 
13,000 
14,000 
14,000 
14,000 
15,000 
20,000 

20,000 
20,000 
30,000 
3,000 
7,000 
14,000 

5,000 
8,000 

Cast  iron  

Gun-metal  cast  iron 
Wrought  iron  
Extra  soft  steel  .  .  . 
Flange  steel  
Firebox  steel 

Machinery  steel.  .  . 
Axle  steel  

Hard   steel    (rail 
steel)  

Tire  steel  
Crucible  (tool)  steel 
Brass,  cast  

Bronze    cast    .    ... 

Tobin  bronze  
Hard-drawn  brass 
wire  

Copper,  rolled  

4  STRENGTH  OF  MATERIALS.  §  7 

11.  In  metals,  a  high  tensile  strength  in  itself  is  no  indi- 
cation of  the  ability  of  the  metal  to  stand  repeated  applica- 
tions of  sudden  stresses,  as  a  high  tensile  strength  usually 
involves   a   lesser   degree  of   ductility  than   is  obtained   in 
metals  of  lower  tensile  strength.     Steel  of  a  tensile  strength 
higher  than  75,000  pounds  is  seldom  used  merely  on  account 
of  its  superior  strength,  but  rather  on  account  of  its  hard- 
ness, which  enables  it  to  better  withstand  abrasion. 

12.  The   loads  given  in  Table    I   are  conservative  safe 
loads  deducted  from  experience  and  observation.     They  are 
given  for  material  free  from  welds  for  such  material  as  can 
be  welded.     While  it  is  possible  to  make  a  weld  as  strong  as 
the  solid  bar,  the  chances  of  the  weld  being  imperfect  are  so 
great  that  it  is  unsafe  to  rely  on  such  a  degree  of  strength  in 
welded    bars.     Furthermore,    the    value  of   the   weld  is  an 
uncertain  quantity  that  cannot  be  determined  by  an  ocular 
inspection  or  any  other  inspection  short  of  actually  pulling 
the  weld  apart  in  a  testing  machine.      Hence,  it  is  advisable 
to  reduce  the  safe  loads  given  in  the  above  table  by  25  per 
cent,   when    a  welded  bar  is  subjected    to  a  tensile  stress. 
When  judging  the  safe  load  of  timber,  due  allowance  must 
be  made  for  knot  holes  and  sappy  spots. 

13.  It  is  often  rather  hard  to  determine  whether  a  stress 
is  steady,  gradually  varying,  or  gradually  applied,  or  sudden. 
For  example,  considering  the   shell  of   a   boiler,   it   would 
appear  on  first  thought  as  though  the  tensile  stress  in  the 
shell  plates  was  a  steady  stress.     But  looking  further  into 
the  problem,  it  will  become  apparent  that  the  stress  varies 
with  the  steam  pressure,  which  gradually  fluctuates  within 
narrow  or  wider  limits.      Hence,  most  designers  would  con- 
sider the  stress  in  a  boiler  shell  as  a  gradually  applied  stress. 
In  a  piston  rod  or  connecting-rod  the  load  is  applied  almost 
instantly  as  soon  as  the  crank  passes  over  the  center,  and, 
hence,    the    stress  would   be   considered    to   be  a    suddenly 
applied  stress.     When  in  doubt,  it  is  good  policy  to  err  on 
the  side  of  safety,  that  is,  to  choose  a  smaller  safe  load  per 
square  inch  of  section. 


§  7  STRENGTH  OF  MATERIALS.  5 

For  special  work,  experience  has  indicated  safe  loads  for 
different  materials  that  fall  below,  or  are  above,  those  given 
in  the  table.  Examples  of  this  will  be  given  later  on. 

14.  The  safe  load  per  square  inch  of  section  is  often 
called  the  working  stress  per  square  Inch,  or  the  "work- 
ing load  per  square  inch.  Care  should  be  taken  not  to 
confound  these  terms  with  working  load,  'Corking  stress, 
safe  load,  or  safe  tensile  strength,  which,  when  applied 
without  the  limitation  as  to  the  unit  of  area,  refer  to  the 
safe  load  the  whole  bar  can  carry. 


RULES  AND  FORMULAS  FOR  TENSILE  STRENGTH. 

15.     Let  W  —  safe  load  in  pounds; 

A  =  area  of  minimum  cross-section ; 
5  =  working  stress  in  pounds  per  square  inch, 
as  given  in  Table  I. 

•  Rule  1. —  The  working  load  in  pounds  for  any  bar  sub- 
jected to  a  tensile  stress  is  equal  to  the  minimum  sectional  area 
of  the  bar,  multiplied  by  the  working  stress  in  pounds  per 
square  inch,  as  given  in  the  table. 

Or,  W=AS. 

EXAMPLE. — A  bar  of  good  wrought  iron  that  is  3  inches  square  is 
to  be  subjected  to  a  steady  tensile  stress;  what  is  the  maximum  load 
that  it  should  carry  ? 

SOLUTION.— According  to  the  table,  a  working  stress  of  13,000  pounds 
may  be  allowed.  Applying  the  rule,  we  have 

W  =  3  X  3  X  13,000  =  117,000  Ib.     Ans. 

Rule  2. —  The  minimum  sectional  area  of  any  bar  sub- 
jected to  a  tensile  stress  should  be  equal  to  the  load  in  pounds, 
divided  by  the  working  stress  in  pounds  per  square  inch,  as 
given  in  the  table. 

Or,  A=- 


6  STRENGTH  OF  MATERIALS.  §  7 

EXAMPLE.—  What  should   be  the  area  of  a  machinery-steel  bar  to 
carry  a  steady  load  of  108,000  pounds  ? 

SOLUTION.  —  According  to  the  table,  a  safe  load  of  15,000  pounds  per 
square  inch  of  section  may  be  allowed.     Then,  applying  the  rule, 

108,000 

A  =  = 


Rule  3.  —  The  working  stress  in  pounds  per  square  inch  is 
equal  to  the  load  in  pounds  divided  by  the  minimum  sectional 
area  of  the  bar. 

W 
Or,  Sf% 

EXAMPLE.  —  A  piston  rod  3  inches  in  diameter  carries  a  piston 
20  inches  in  diameter.  With  a  steam  pressure  of  100  pounds  to  the 
square  inch,  what  is  the  stress  per  square  inch  of  section  of  the  rod  ? 

SOLUTION.—  The  load  on  the  piston  is  202  X  .7854  X  100  =  31,416 
pounds.  The  area  of  the  rod  is  3*  X  .7854  =  7.0686  square  inches. 
Then,  applying  the  rule  just  given,  we  have 

~      31,416 
=  7Q686  =  4,444.4  Ib.     Ans. 


CHAINS. 

16.  Chains  made  of  the  same  size  iron  vary  in  strength, 
owing  to  the  different  kinds  of   links  from  which  they  are 
made.     It  is  a  good  practice  to  anneal  old  chains  that  have 
become  brittle    by  overstraining.     This    renders   them  less 
liable  to  snap  from  sudden  jerks.     It  reduces  their  tensile 
strength,  but  increases  their  toughness  and  ductility,  which 
are  sometimes  more  important  qualities. 

When  annealing,  care  should  be  taken  that  a  sufficient  heat 
be  applied,  otherwise  no  benefit  will  be  gained;  the  chains 
ought  to  be  heated  to  a  cherry  red,  say  1,300°  F.,  at  least. 

17.  The  loads  that  can  safely  be  lifted  with  chains  are 
given  in  the  following  table.     The  safe  load  given  is  one- 
quarter  of  the  steady  load  at  which  they  may  be  expected 
to  break.     The  table  has  been  deduced  from  one  published 
by  the  Lukens  Iron  and  Steel  Company. 


7  STRENGTH  OF  MATERIALS. 

TABLE  II. 


STRENGTH  OF  CHAINS. 


Diameter  of 
Rod  of  Which 
Link  Is  Made. 

Safe  Load. 
Pounds. 

Diameter  of 
Rod  of  Which 
Link  Is  Made. 

Safe  Load. 
Pounds. 

A 

430 

1 

12,500 

1 

770 

*t 

15,000 

yV 

1,200 

H 

18,000 

t 

1,750 

If 

22,000 

yV 

2,350 

if 

26,000 

1 

3,100 

if 

31,000 

TV 

4,000 

if 

36,000 

1 

4,800 

if 

41,000 

tt 

5,800 

2 

47,000 

t 

6,900 

^ 

56,000 

it 

8,000 

** 

69,000 

1 

9,400 

2f 

84,000 

if 

11,000 

3 

100,000 

STRENGTH  OF  ROPES. 


HEMP  ROPES. 

18.  The  strength  of  hemp  ropes  does  not  depend  entirely 
on  the  quality  of  the  material  and  the  cross-section  of  the 
rope,  but  to  a  large  extent  on  the  method  of  manufacture 
and  the  amount  of  twisting.  A  hemp  rope  is  made  up  of 
fibers  twisted  together  to  form  a  yarn,  several  of  which  are 
laid  up  left-handed  into  strands,  which  in  turn  are  twisted 
up  right-handed  into  what  is  known  as  plain-laid  rope. 

Cable-laid  or  hawser-laid  rope  is  left-handed  rope  of 
nine  (9)  strands. 

Shroud-laid  rope  is  formed  by  adding  another  strand  to 
a  plain-laid  rope.  Since  the  four  strands  on  application  of 


8 


STRENGTH  OF  MATERIALS. 


strain  would  sink  in  and  detract  from  the  rope's  strength 
by  an  unequal  distribution  of  strain,  they  are  laid  up  around 
a  heart,  which  is  a  small  rope,  made  soft  and  elastic,  and 
about  one-third  the  size  of  the  strands. 

19.  Ordinary  hemp  rope  is  designated  by  its  circumfer- 
ence; rope  for  transmission  purposes,  which  is  either  iron, 
steel,  or  manila  rope,  is  designated  by  its  diameter.  It  is 
well  to  keep  this  in  mind.  The  loads  that  can  safely  be 
hoisted  with  hemp  ropes  of  different  sizes  is  given  in  the 
following  table  prepared  by  Ensign  F.  R.  Brainard,  United 
States  Navy. 

TABLE  III. 


STRENGTH  OF  HEMP  ROPES. 


Circumference. 
-Inches. 

Safe  Load. 
Pounds. 

Circumference. 
Inches. 

Safe  Load. 
Pounds. 

1.00 

65 

5.75 

2,680 

1.25 

125 

6.00 

2,975 

1.50 

185 

6.50 

3,470 

1.75 

255 

7.00 

3,965 

2.00 

320 

7.50 

4,570 

2.25 

410 

8.00 

5,175 

2.50 

500 

8.50 

5,910 

2.75 

610 

9.00 

6,640 

3.00 

715 

9.50 

'7,370 

3.25 

850 

10.00 

8,105 

3.50 

980 

10.50 

8,970 

3.75 

1,135 

11.00 

9,840 

4.00 

1,285 

11.50 

11,015 

4.25 

1,450 

12.00 

12,190 

4.50 

1,610 

12.50 

13,365 

4.75 

1,750 

13.00 

14,540 

5.00 

1,990 

13.50 

15,845 

5.25 

2,190 

14.00 

17,150 

5.50 

2,385 

14.50 

18,450 

§  7  STRENGTH  OF  MATERIALS.  9 

20.  The  above  table  applies  to  new  ropes  or  ropes   in 
first-class  condition.      If  the  rope  is  used  for  a   block  and 
tackle,  the  bending  of  the  rope  around  the  pulleys  will  cause 
a  rapid  deterioration  of  the  inside,  owing  to  the  chafing  and 
sliding  of  the  strands  and  yarns  upon  one  another.     The 
outside  appearance  of  a  rope  used  for  block  and  tackle  is  in 
itself  no  indication  of  its  quality ;  ropes  are  frequently  found 
that  appear  perfectly   sound  when  judged  by  their  outside 
appearance   and    yet  are   entirely   unsafe.     To   inspect  the 
rope,   take  it  in   both  hands  and  untwist  it  sufficiently   to 
expose    the    inner  surfaces  that  have  been  chafing  against 
one  another.      If  a  considerable  number  of  broken  fibers  are 
found,  the  safe  load  should  be  reduced  by  50  per  cent.      If  a 
rather  large  quantity  of  the   fibers  have   been   reduced   to 
powder,    the    rope    should    be    condemned    and    preferably 
destroyed  immediately,  in  order  to  remove  any  temptation 
to  use  it.     A  rope  in  this  condition  is  liable  to  give  way 
suddenly  under  a  very  light  load. 

Ropes  used  for  slings  and  lashings,  as  a  general  rule,  are 
ruined  by  the  external  chafing  they  receive  and,  hence, 
their  safety  can  be  determined  by  their  outward  appearance. 

21.  Slings   are    chiefly   used    for    attaching   tackles   to 
machine  parts  during  the  erection  and  repair  of  machinery. 
They  are  made  by  taking  a  piece  of  rope  a  little  more  than 
twice  the  required  length  of  the  rope  and  joining  the  ends  by 
splicing,  the  splice  known  as  a  short  splice  being  usually 
employed.      It    should    be   distinctly    understood    that    this 
splice  is  not  intended  nor  adapted  for  ropes  that  are  to  pass 
over  a  pulley. 

22.  Splicing  a  Sling. — To  make  a  short  splice  in  a  sling, 
untwist  the  strands  for  some  distance  from  each  end  of  the 
rope.     Take  an  end  in  each  hand  and  lay  the  strands  within 
one  another,  as  shown  in  Fig.   1   (a).     The  strands  1',  2', 
and  3'  may  be  tied  to  the  left-hand  end  for  convenience  in 
handling.      Now  take  the  strand  1,  pass  it  over  strand  2' 
and  under  strand  ./',  as  shown  in  Fig.  1   (b] ;  then,  draw  it 


10 


STRENGTH  OF  MATERIALS. 


tight.      Next,   take  strand  2,  pass  it  over  3'  and  under  2', 
drawing  it  up  tightly.      Now  take  strand  <?,  pass  it  over  1' 

and  under  3';  then 
draw  it  up.  The 
splice  will  now  appear 
as  shown  in  Fig.  1  (r). 
Weave  the  strands  1', 
2',  and  3'  into  the  left- 
hand  rope  in  the  same 
manner.  Continue 
weaving  the  right- 
hand  and  left-hand 
strands  alternately 
until  the  length  of  the 
splice  is  about  6  times 
the  diameter  of  the 
rope.  Draw  the 
FIG-  *•  strands  tight  and  sub- 

ject the  splice  to  a  strain.     Then  cut  off   the  projecting 
ends  of  the  strands. 


23.  Use  of  Slings. — The  weight  *  that  can  safely  be 
lifted  by  a  sling  depends  not  only  on  the  size  of  the  rope  of 
which  it  is  made,  but  also  on  the  manner  in  which  it  is 
attached  to  the  hook  of  the  tackle  block.  Thus,  in  Fig.  2, 
the  sling  is  simply  hooked  over  the  hook. 

Owing  to  the  sharp  bend  over  the  hook,  the  strength  of 
the  sling  is  greatly  reduced,  and  the  direct  load  for  this 
manner  of  attachment  should  not  exceed  the  safe  load  given 
in  the  table  for  one  rope  of  equal  circumference,  although 
the  load  is  borne  by  two  parts  of  the  rope. 

When  the  sling  is  attached  to  the  hook  by  doubling  over, 
as  shown  in  Fig.  3,  its  greatest  strength  is  realized,  and  for. 
the  two  parts  of  the  rope  parallel  to  each  other,  the  stress 
in  the  rope  will  be  equal  to  half  the  load,  as  the  load  is  sup- 
ported by  two  ropes.  Then,  with  a  sling  doubled  over  as 
shown,  a  load  twice  as  great  can  safely  be  hoisted  than 
when  applied  as  shown  in  Fig.  2. 


STRENGTH  OF  MATERIALS 


11 


In  the  foregoing  statement,  it  is  assumed  that  the  sling  is 
fastened  to  the  work  without  a  sharp  bend,  the  equivalent 
of  a  single  loop  over  a  hook.  The  stress  in  the  two  parts 
of  a  sling  will  be  least  when  they  are  parallel  to  each  other; 
hence,  a  sling  will  then  lift  the  greatest  load  with  safety. 
The  stress  becomes  greater  when  the  angle  between  the 
two  parts  of  the  sling  is  increased;  its  magnitude  for  any 
angle  can  be  determined  by  the  method  of  the  triangle  of 
forces. 

24.  A  quick  way  of  attaching  a  rope  to  a  hook  is  that 
shown  in  Fig.  4,  and  known  as  a  Blackwall  hitch.  When 


making  this  hitch,  always  make  it  as  shown  in  the  figure; 
that  is,  let  the  detached  end  lie  between  the  hook  and  the 
end  sustaining  the  load. 

25.  The  proper  size  of  a  rope  for  a  block  and  tackle  is 
fixed  by  the  size  of  the  groove  turned  in  the  pulleys.  Then, 
from  the  table,  take  the  safe  working  load  of  the  rope,  and 
using  this  as  the  force  applied  to  the  free  end  of  the  rope, 
the  maximum  load  a  given  block  and  tackle  can  lift  safely 
can  be  calculated  by  the  rule  given  in  connection  with  the 
subject  of  pulleys. 


12  STRENGTH  OF  MATERIALS.  §  7 

MANILA  ROPE. 

26.  Manila  rope  is  used  chiefly  for  hoisting  purposes 
and    for   power  transmission.      The  so-called   "Stevedore" 
manila  rope  has  the  fibers  lubricated  with  a  mixture  of  tallow 
and  plumbago  during  the  process  of    manufacture.      This 
lubrication  prevents  internal  chafing  and  wear  to  a  large 
extent. 

Manila  rope  for  hoisting  and  power-transmission  purposes 
is  usually  made  with  four  strands,  that  is,  shroud  laid. 
Hoisting  rope  is  ordered  by  circumference ;  transmission 
rope  by  diameter. 

27.  The  working  stress  on  manila  hoisting  rope  should 
not  exceed  that  given  in  the  table  below,   which  was  pub- 
lished by  C.  W.  Hunt  &  Company. 

TABLE  IT. 


STRENGTH  OF  MANILA  HOISTING  ROPE. 


Circumference  of  Rope. 
Inches. 

Working  Stress. 

3 

350 

8* 

500 

4 

650 

*4 

800 

5 

..    1,000 

Under  ordinary  conditions,  with  a  working  stress  not 
exceeding  that  given  in  the  table,  a  manila  hoisting  rope 
will  last  until  a  total  load  of  about  6,000  tons  has  been 
hoisted  with  it.  Under  exceptionally  favorable  circum- 
stances, where  the  rope  runs  over  large  sheaves,  a  manila 
rope  has  hoisted  20,000  tons.  Manila  ropes  for  hoisting 
purposes  should  not  be  spliced,  as  it  is  difficult  to  make  a 
splice  that  will  not  pull  out  while  running  over  the  sheaves, 


§  7  STRENGTH  OF  MATERIALS.  13 

and  the  increased  wear  to  be  obtained  from  splicing  a  broken 
hoisting  rope  is  usually  very  small. 

28.  Manila  rope  for  power  transmission  is  made  from 
^  inch  up  to  2  inches  in  diameter.  The  greatest  horsepower 
can  be  transmitted  by  a  rope  when  the  rope  speed  is  about 
5,000  feet  per  minute.  When  run  at  greater  speed,  the 
effect  of  the  centrifugal  force  in  the  rope  rapidly  decreases 
the  horsepower  that  can  be  transmitted.  The  size  of  the 
pulley  has  an  important  bearing  on  the  wear  of  the  rope; 
in  general,  it  should  not  be  smaller  than  40  times  the  diam- 
eter of  the  rope,  and  as  much  larger  as  it  can  conveniently 
be  made. 


WIRE   ROPES. 

29.  Wire  rope  is  made  of  iron  and  steel  wire.     It  is 
stronger  than  hemp  rope,  and,  to  carry  the  same  load,  is  of 
smaller  diameter. 

In  substituting  steel  for  iron  rope,  the  object  in  view 
should  be  to  gain  an  increase  of  wear  from  the  rope,  rather 
than  to  reduce  the  size.  A  steel  rope  to  be  serviceable 
should  be  of  the  best  obtainable  quality,  because  ropes  made 
from  low  grades  of  steel  are  inferior  to  good  iron  ropes. 

30.  Rules  foi-  the  strength  of  wire  ropes  : 

Let  W  =  maximum  of  working  load  in  pounds; 
C  =  circumference  of  rope  in  inches. 

Rule  4. —  Tlie  maximum  working  load  in  pounds  that 
sliould  be  allo^cved  on  any  iron-wire  rope  is  equal  to  the  square 
of  the  circumference  of  the  rope  in  inches,  multiplied  by  600. 

Or,  W  =  600  C\ 

EXAMPLE. — What  is  the  maximum  load  in  pounds  that  should  be 
carried  by  an  iron  rope  whose  circumference  is  4^  inches  ? 

SOLUTION. — Applying  the  rule  just  given,  we  have 
W  =  600  X  4.52  =  12,150  Ib.     Ans. 


14  STRENGTH  OF  MATERIALS.  §  7 

Rule  5. —  The  circumference  of  any  iron-wire  rope  in  indies 
is  equal  to  the  square  root  of  the  maximum  working  load  in 
pounds,  multiplied  by  .0408. 

Or,  C=.0408/F: 

EXAMPLE. — A  maximum  working  load  of  12,150  pounds  is  to  be 
carried  by  an  iron-wire  rope ;  what  should  be  the  minimum  circumfer- 
ence of  the  rope  ? 

SOLUTION. — Applying  rule  5,  we  have 

C  =  .0408  |/12,150  =  4£  in.     Ans. 

Rule  6. —  The  above  rules  and  formulas  are  also  made 
applicable  when  computing  the  safe  strength  of  steel-wire 
rope  by  substituting  the  constant  1,000  for  the  constant  600, 
anal  .0316  for  .0408. 

EXAMPLE. — What  is  the  maximum  load  in  pounds  that  should  be 
carried  by  a  steel-wire  rope,  the  circumference  of  which  is  4£  inches  ? 
SOLUTION.— Applying  rule  4,  we  have 

W  -  1,000  X  4.52  =  20,250  Ib.     Ans. 

EXAMPLE.— A  maximum  working  load  of  10,485  pounds  is  to  be 
carried  by  a  steel-wire  rope ;  what  should  be  the  minimum  circumfer- 
ence of  the  rope  ? 

SOLUTION. — Applying  rule  5,  we  have 

C  =  .0316  4/10,485  =  3.24  in.     Ans. 


EXAMPLES  FOB  PRACTICE. 

1.  What  should  be  the  diameter  of  a  machinery-steel  piston  rod  of 
a  steam  engine  to  resist  tension,  if  the  piston  is  19  inches  in  diameter 
and  the  pressure  is  85  pounds  per  square  inffh  ?        Ans.  2£  in.,  nearly. 

2.  What  safe  load  will  a  cast-iron  bar  of  rectangular  cross-section 
7£  in.  by  3£  in.  support  if  subjected  to  shocks  ?    The  bar  is  in  tension. 

Ans.  42,000  Ib. 

3.  What  is  the  stress  per  square  inch  on  a  piece  of  timber  8  inches 
square  that  is  subjected  to  a  steady  pull  of  60,000  pounds  ? 

Ans.  937.5  Ib.  per  sq.  in. 

4.  What  should  be  the  allowable  working  load  for  a  steel-wire  rope 
whose  circumference  is  3J  inches  ?  Ans.  14,062.5  Ib. 

5.  What  should  be  the  circumference  of  an  iron-wire  rope  to  sup- 
port a  load  of  20,000  pounds  ?  Ans.  5f  in.,  nearly. 


STRENGTH  OF  MATERIALS. 


15 


CRUSHING  STRENGTH  OF    MATERIALS. 

31.  The    crushing   strength,   of    any    material    is    the 
resistance  offered  by  its  fibers  to  being  pushed  together. 

32.  To  obtain  only  compression,   the    length    of   a   rod 
should  not  be  more  than  five  times  greater  than  its  least 
diameter  or  its  least  thickness  when  it  is  a  rectangular  rod. 
Such  a  rod  is  called  a  sliort  column. 

If  a  bar  is  long  compared  with  its  cross  dimensions,  the 
load,  if  sufficiently  great,  will  cause  it  to  bend  sidewise 
under  the  compressive  force,  and  we  have,  then,  not  only 
compression,  but  compression  compounded  with  bending. 

33.  When  a  bar  or  rod  is  longer  than  five  times  its  least 
diameter  or  least  thickness,  it  is  called  a  long  column,  or 
long  pillar. 

34.  The  shape  of  the  ends  of  a  column  has  great  influ- 
ence on  its  strength.      In  Fig.  o  are  shown  three  columns 
with  differently  shaped  ends.  W^^M^M^M^^^M^ 

It  has  been  proved  by  the  aid  %, 
of  higher  mathematics  that,  theo-  * 
retically,   a    pillar,   as   a,   having 
flat  or  fixed  ends,   is  4  times  as 
strong  as  one  that  has  round  or 
movable  ends,  as  the  pillar  c,  and 
1|  times  as  strong  as  one  having 
one  flat  and    one    round  end,   as 
the  pillar  /;;  b  is  thus  2£  times  as 
strong   as   c.     It    has  .also   been  •% 

found  that  if  three  pillars  «,  b,  c,  ^,..:, ..,....,..,, , /;,, 

which    have  the  same    cross-sec-  FlG- 5- 

tion,  are  to  carry  the  same  load  and  be  of  equal  strength, 
their  lengths  must  be  as  the  numbers  2,  1£,  and  1,  respect- 
ively. 

In  practice,  however,  the  ends  of  the  pillars  b  and  c  are 
not  generally  made  as  shown  by  the  figure,  but  have  holes 
at  their  ends  into  which  pins  are  fitted  that  are  fastened  to 
some  other  piece.  In  such  cases,  it  has  been  found  that  a  is 


16 


STRENGTH  OF  MATERIALS. 


2  times  as  strong  as  c  and  that  b  is  \\  times  as  strong  as  c. 
That  is,  in  actual  practice,  a  column  fixed  as  at  c  is  really 
\  as  strong  as  one  fixed  as  at  a,  instead  of  being  only  $  as 
strong,  as  given  above. 

An  example  of  a  column  fixed  at  both  ends  is  the  piston 
rod  of  a  steam  engine.  The  valve  stem  of  a  slide-valve 
engine  may  be  considered  as  an  example  of  a  column  fixed  at 
one  end  and  movable  at  the  other.  The  connecting-rod  of 
a  steam  engine  is  a  good  example  of  a  column  having  two 
movable  ends. 

35.  Average  values  of  the  crushing  strength  of  various 
materials  used  in  engineering  are  given  in  the  table  below. 

TABLE  V. 


CRUSHING   STRENGTH  OF  MATERIALS. 


Material. 

Crushing 
Strength. 
Pounds  per 
Sq.  In. 

Safe  Loads.       Pounds. 

Sudden. 

Gradual. 

Steady. 

Cast  iron  

80,000 
45,000 
60,000 
9,000 
7,000 
3,000 
6,000 
2,000 

7,000 
3,500 
4,000 
800 
600 
250 

11,000 
4,500 
6,000 
1,200 
850 
380 

14,000 
6,000 
8,000 
1,500 
1,100 
500 
200 
100 

Wrought  iron  
Machinery  steel  
Cast  brass 

Timber   dry  ; 

Timber,  wet  

Stone 

Brick  

STRENGTH  OF  COLUMNS. 

36.  In  practice,  columns  subjected  to  a  compressive 
stress  are  made  of  cast  iron,  wrought  iron,  steel,  or  timber. 
For  short  columns,  the  area,  etc.  can  be  calculated  by  the 
same  rules  that  have  been  given  for  tensile  stresses,  substi- 
tuting the  working  loads  per  square  inch  given  in  the  above 
table. 


§  7  STRENGTH  OF  MATERIALS.  17 

37.  The  safe  working  load  on  long  columns  is  given 
by  the  following  rule,  which  is  applicable  to  columns  the 
length  of  which  does  not  exceed  40  times  their  least  diameter 
or  their  least  thickness  when  rectangular.  The  rule  given 
applies  to  columns  uniform  throughout  their  length.  The 
rules  for  tapering  columns  involve  considerable  mathemati- 
cal knowledge  and  are  so  rarely  used  in  practice  that  they 
will  not  be  given  here. 

Let      C  =  safe  crushing  load,  as  given  in  the  table; 
5"  =  sectional  area  in  square  inches; 
L  =  length  of  column  in  inches ; 
d  =  least  thickness  of  rectangular  column,  or  diam- 
eter of  round  column,  or  dimension  indicated 
by  the  arrowheads  in  the  following  table,  in 
inches ; 

W '=  safe  working  load  in  pounds; 
A  =  area  of  the  two  flanges  in  square  inches ; 
B  =  area  of  the  web  in  square  inches; 
a  =  constant  corresponding  to  the  cross-section  of 
the  column,  as  given  in  Tables  VI,  VII,  and 
VIII. 

Rule  7. —  The  safe  working  load  of  a  long  column,  in 
pounds,  is  equal  to  the  safe  working  load  corresponding  to  the 
kind  of  stress  given  in  Table  V  multiplied  by  the  sectional 
area  of  the  column,  in  square  inches,  and  the  product  divided 
by  1  plus  the  quotient  obtained,  by  dividing  the  square  of  the 
length  of  the  pillar  in  inches  by  the  square  of  the  diameter  (or 
least  thickness,  if  rectangular]  multiplied  by  the  value  of  a. 

Or.  "'         CS 


38.  When  using  this  formula,  first  obtain  the  value  of  C 
from  Table  V.  Next,  calculate  the  area  of  the  cross-section 
of  the  pillar.  Then  find  the  value  of  a  from  one  of  the  last 
three  tables.  Finally,  be  sure  that  the  length  of  the  column 
has  been  reduced  to  inches  before  substituting  in  the  for- 
mula. 

//.     8.    I.— 22 


18 


STRENGTH  OF  MATERIALS. 


TABLE  VI. 


CONSTANTS  FOR  WROUGHT-IRON  AND  STRUCTURAL- 
STEEL,  PILLARS. 


^ross-Section  of  Pillar. 

When  Both 
Ends  of  the 
Pillar  Are 
Flat  or  Fixed. 

When  One 
End  of  the 
Pillar  Is  Flat 
or  Fixed 
and  the  Other 
Round  or 
Movable. 

When  Both 
Ends  of  the 
Pillar  Are 
Round  or 
Movable. 

^P       Round. 
h-«H 

2,250 

1,500 

1,125 

1               Square  or 
j    Rectangle. 

3,000 

2,000 

1,500 

nl  nin  bquare 
Tube. 

6,000 

4,000 

3,000 

*2tj       Thin  Round 
\J       Tube- 

•vmnmi       Angle  With 
Equal  Sides. 

jj~1       Cross  With 
^f"       Equal  Arms. 

I-«W 
T<F          T  "Ream 

4,500 
1,500 

1,500 
A 

3,000 
1,000 

1,000 

A 

2,250 

750 
750 
1    ^00  v 

A  -f-  B 

A+B 

STRENGTH  OF  MATERIALS. 


19 


TABLE  Til. 


CONSTANTS  FOR  CAST-IRON  PIL.L.ARS. 


When  One 

Cross-Section  of  Pillar. 

When  Both 
Ends  of  the 
Pillar  Are 
Flat  or  Fixed. 

End  of  the 
Pillar  Is  Flat 
or  Fixed 
and  the  Other 
Round  or 

When  Both 
Ends  of  the 
Pillar  Are 
Round  or 
Movable. 

Movable. 

(       \       Round. 

281.25 

187.5 

140.625 

{.dJ    I--**-1) 

ill              Square  or 
\    Rectangle. 

375.00 

250.0 

187.500 

Thin  Square 
Tube. 

750.00 

500.0 

375.000 

JAJ       Thin  Round 

w    Tube- 

562.50 

375.0 

281.250 

JPBB       Angle  With 
Equal  Sides. 

187.50 

125.0 

93.750 

Cross  With 
^^       Equal  Arms. 

187.50 

125.0 

93.750 

tfj 

A 

A 

. 

*PP^        I  Bccim 

375  X 

°*50  V 

105  v 

/Ct)U   A       *      .       /> 

A  -(-  b 

L/CO    /\    ~i      j       ^ 

20 


STRENGTH  OF  MATERIALS.  §  7 

TABLE  Till. 


CONSTANTS  FOR  WOODEN  PILLARS. 


When  One 

Cross-Section  of  Pillar. 

When  Both 
Ends  of  the 
Pillar  Are 
Flat  or  Fixed. 

End  of  the 
Pillar  Is  Flat 
or  Fixed 
and  the  Other 
Round  or 

When  Both 
Ends  of  the 
Pillar  Are 
Round  or 
Movable. 

Movable. 

Round. 

187.5 

125.00 

93.75 

M<|   |"-<M   Square 

ill  iilli)  or  Rect- 

250.0 

166.66 

125.00 

H              angle. 

K-rfH       Hollow 

IF*8!       Square  Made 

500.0 

333.33 

250.00 

JLaJI       of  Boards. 

To  find  the  proper  value  of  a  in  any  example,  first  turn  to 
the  table  dealing  with  the  material  in  question,  and  find  the 
figure  corresponding  to  the  given  cross-section ;  in  the  hori- 
zontal line  containing  this  are  three  numbers  corresponding 
to  the  different  conditions  of  the  ends  of  the  column.  From 
these  numbers  select  the  one  corresponding  to  the  given 
conditions  of  the  column  to  be  calculated,  and  this  will  be 
the  required  value  of  a. 

NOTE. — If  the  length  of  the  pillar  is  given  in  feet,  be  sure  to  reduce 
it  to  inches  before  substituting  in  the  formula. 

39.  Rule  7  cannot  be  transformed  to  determine  directly 
the  diameter  or  area  of  a  column  of  a  given  cross-section 
required  to  sustain  a  stated  load.  An  area  and  a  corre- 
sponding value  of  d  must  be  assumed  and  the  safe  load 


§  7  STRENGTH  OF  MATERIALS.  21 

corresponding  to  these  values  calculated.  If  the  safe  load  is 
smaller  than  the  load  to  be  sustained,  a  larger  area  and 
value  of  d  must  be  chosen.  If  the  calculated  safe  load  is 
larger,  a  somewhat  smaller  area  and  value  of  d  may  be 
tried. 

When  applying  the  rule  given  to  a  column  subject  to 
wear,  as,  for  instance,  the  piston  rod  of  an  engine,  the  diam- 
eter of  the  rod  should  be  increased  somewhat  over  that  given 
by  calculation  in  order  to  allow  for  subsequent  truing  up. 
This  allowance  may  be  from  £  to  \  of  an  inch,  according  to 
the  judgment  of  the  designer. 

4O.  A  rough-and-ready  rule  for  the  size  of  a  piston 
rod  is  to  make  it  one-sixth  the  diameter  of  the  cylinder 
of  a  simple  engine.  This  rule  of  thumb  allows  us  to  choose 
a  value  of  5  and  d  that  will  answer  very  well  indeed  for 
the  first  trial. 

ILLUSTRATION.— What  size  machinery-steel  piston  rod  is  required 
for  a  simple  engine  having  a  piston  36  inches  in  diameter  to  carry  a 
steam  pressure  (gauge)  of  100  pounds  per  square  inch;  the  rod  is 
70  inches  long  ? 

According  to  the  rule  given,  an  approximate  size  of  the  piston  rod 
will  be  36  -*-  6  =  6  inches.  The  area  corresponding  to  this  is  62  X  .7854 
=  28.27  square  inches.  From  Table  VI,  a  =  2,250.  The  piston  rod 
being  subjected  to  suddenly  applied  stresses,  by  Table  V,  the  allowable 
safe  working  load  per  square  inch  is  4,000  pounds.  The  working  load 
on  the  piston  rod  is  36'2  X  -7854  X  100  =  101,787.8  pounds.  Applying 
rule  7  and  substituting  values,  we  get 

106,629  pounds, 


2,250  X  6* 


which  is  a  safe  load  on  a  6-inch  piston  rod  under  the  given  conditions. 
This  tends  to  show  that  the  rod  is  rather  large. 

Now  try  a  51-inch  rod.  The  area  of  the  rod  is  5£2  X  .7854  =  27.1 
square  inches,  nearly. 

Applying  rule  7  again  and  substituting  the  new  values  of  S  and  d, 
we  get 

TZr        4,000  X  27.1 

TO1 =  102,264  pounds,  nearly. 


1  + 


2,250  X  54* 


22  STRENGTH  OF  MATERIALS.  §  7 

It  is  thus  seen  that  a  5J-inch  rod  is  just  about  right.  Most  designers 
would  allow  -J-  inch  for  truing  up,  thus  making  the  rod  6  inches  in 
diameter. 

In  this  particular  case  the  size  of  rod  found  by  calculation 
agrees  with  that  given  by  the  rough-and-ready  rule.  This 
cannot  be  in  every  case,  however,  as  will  readily  be  seen 
if  a  steam  pressure  of  150  pounds  be  substituted  for 
100  pounds  and  the  rod  be  calculated  for  this  pressure.  It 
will  then  be  found  that  a  rod  larger  than  6  inches  will  be 
required. 

41.  In  calculating  the  size  of  piston  rods  in  actual 
work,  it  is  good  practice  to  assume  at  least  a  steam  pres- 
sure of  100  pounds  for  the  calculation,  even  though  it  is 
intended  to  carry  a  lower  pressure.  An  extra  margin  of 
strength  is  thus  provided  which  may  be  needed  in  the 
future,  as  cases  are  very  frequent  where  engines  designed 
for  75  pounds  have  ultimately  been  run  at  100  pounds 
pressure. 

4:2.  In  engineers'  examinations,  candidates  for  license 
are  often  asked  to  calculate  the  size  of  the  piston  rod  for  a 
given  engine.  A  rule  that  is  easily  remembered  is  the  fol- 
lowing, which  assumes  the  piston  rod  to  be  a  short  column 
and  makes  allowance  for  failure  by  bending  by  reducing  the 
allowable  working  stress  given  in  the  table  by  10  per  cent., 
thus  making  the  safe  working  stress  for  steel  piston  rods 
3,600  pounds  per  square  inch  and,  say,  3,100  pounds  for 
iron. 

This  rule  will  give  fairly  good  results  within  the  limits  of 
ordinary  practice  and  will  usually  satisfy  an  examiner.  For 
an  actual  design,  however,  rule  7,  which  is  more  rational,  is 
preferable. 

Let  vS  =  area  of  piston  rod ; 

/  =  load    in    pounds   on    the    rod  =  area   of    piston 

X  steam  pressure ; 
c  =  safe  working  load  in  pounds. 


§  7  STRENGTH  OF  MATERIALS.  23 

Rule  8.  —  Multiply  the  area  of  the  piston  by  tJie  steam  pres- 
sure and  divide  the  product  by  the  safe  working  load  corre- 
sponding to  the  material.  The  quotient  will  be  the  required 
area. 

Or,  S  =  ~. 

EXAMPLE.  —  What  size  steel  piston  rod  is  required  for  a  36-inch 
cylinder  with  a  steam  pressure  of  100  pounds  ? 

SOLUTION.  —  The  safe  working  load  to  be  used  in  applying  rule  8  is 
3,600  pounds. 

362  X  .7854  X  100 
Then,          S=—        .  —  =.  28.27  square  inches. 


The  corresponding  diameter  is  |/     _      =  6  in.     Ans 


EXAMPLES  FOR  PRACTICE. 

1.  A  round  wrought-iron  column  4  inches  in  diameter  and  60  inches 
long  is  subjected  to  a  steady  load.     The  column  is  fixed  at  one  end 
and  movable  at  the  other.     What  load  will  it  sustain  ? 

Ans.  65,564  lb.,  nearly. 

2.  A  solid    machinery-steel    column   with   both    ends    hinged    is 
4i  inches  in  diameter  and  10  feet  long.     It  is  subjected  to  a  suddenly 
applied  stress.     What  is  the  safe  working  load  ?  Ans.  38,979  lb. 

3.  A  rectangular  wooden  column  is  14  feet  long.     One  end  is  fixed 
and  one  end  is  movable.     If  the  cross-section  is  12  in.  X  8  in.,  what  is 
its  safe  load  for  a  steady  stress  ?  Ans.  28,963  lb. 


TRANSVERSE   STRENGTH  OF 
MATERIALS. 

43.  The  transverse  strength  of  any  material  is  the 
resistance  offered  by  its  fibers  to  being  broken  by  bending. 
As,  for  example,  when  a  beam,  bar,  rod,  etc.  which  is  sup- 
ported at  its  ends  is  broken  by  a  force  applied  between  the 
supports. 

The  transverse  strength  of  any  beam,  bar,  rod,  etc.  is 
proportional  to  the  product  of  the  square  of  its  depth 


24  STRENGTH  OF  MATERIALS.  §  7 

multiplied  by  its  width;  consequently,  it  is  more  economi- 
cal to  increase  the  depth  than  the  width. 

44.  The  safe  load  that  can  be  carried  by  beams,  bars, 
rods,  etc.  of  uniform  cross-section  depends  on  the  material, 
the  manner  in  which  the  beam  is  supported  and  loaded,  and 
on  the  shape  of  the  cross-section.     A  beam  that  is  rigidly 
fixed  at  one  end  and  free  at  the  other,  when  subjected  to 
a  transverse  stress,  is  called  a  cantilever. 

45.  The  safe  working  loads  on  beams    supported    and 
loaded  in  different  ways  can  be  found  by  applying  the  cor- 
responding formula  given  in  Table  IX.      In  the  formulas 
given,   W=  the  safe  load  in  pounds,  and  /  =  length  of  beam 
in  inches  to  be  taken  as  shown  in  the  illustrations  of  Table  IX. 
To  apply  the  formulas  to  a  beam,  the  values  of  5  must  be 
taken  from  Table  X ;  and  the  value  of  R  is  to  be  computed 
by  the  formulas  given  in  Table  XI.      It  may  then  be  substi- 
tuted.    Attention  is  called  to  the  fact  that  it  is  absolutely 
necessary  to  reduce  the  length  of  the  beam  to  inches ;  fur- 
thermore,  the  load    W  on  a  uniformly  loaded  beam  is  the 
total  load  for  the  length  /,  and  not  the  load  per  unit   of 
length. 

In  the  table  giving  the  value  of  R,  the  letter  A  denotes 
the  whole  area  of  the  section  in  square  inches,  that  is,  taking 
the  hollow  cylinder,  for  example,  it  denotes  the  area  corre- 
sponding to  its  outside  diameter.  The  letter  a  stands  for 
the  area  of  the  hollow  part  of  the  section.  In  the  formulas 
where  A  is  applied  to  a  solid  section,  it  stands  for  the  area 
of  the  section. 

The  values  of  the  constant  5  have  been  determined  from 
practical  experience  and  are  safe,  conservative  values  for  the 
conditions  stated. 

EXAMPLE  1. — A  cast-iron,  solid,  rectangular  beam  8  inches  deep  and 
4  inches  wide  rests  upon  two  supports  8  feet  apart.  What  steady  load 
will  it  safely  support  when  the  load  is  applied  at  the  middle  ? 

SOLUTION. — According  to  Table  XI,  the  formula  to  be  used  is 
W '=  — -. — .  For  a  steady  stress  on  cast  iron,  Table  X  gives  7,500  as 


TABLE  IX. 


FORMULAS  FOR  STRENGTH  OF  BEAMS. 


Manner  of  Supporting  Beams. 


Remarks. 


Formula  for 

Safe  Load. 

Pounds. 


Cantilever    load    W   at 
free  end. 


Cantilever,    uniformly 
loaded.    W '=  total  load. 


Simple  beam  resting  on 
two  supports,  load  W 
at  middle. 


Simple  beam  resting  on 
two  supports,  uni- 
formly loaded.  W 
=  total  load  on  length  /. 


Simple  beam  resting  on 
two  supports,  single 
load  W  not  in  the  mid- 
dle. 


Beam  rigidly  fixed  at 
both  ends,  load  W  in 
middle. 


Beam  rigidly  fixed  at 
both  ends,  uniformly 
loaded.  W—  total  load 
on  length  /. 


SR 

~T 


2SR 


4SR 


W  = 


ZSR 


SR 


8SR 


12  SR 


STRENGTH  OF  MATERIALS.  §  7 

TABLE  X. 


VALUES  OF  S. 


Nature  of  Load. 


Material. 

Suddenly 
Applied. 

Gradually 
Applied. 

Steady. 

Cast  iron.N  

2,250 

3,000 

7,500 

Wrought  iron 

4  000 

6  000 

13  700 

Structural  steel  
Brass 

5,000 
1  100 

7,500 
1  500 

16,800 
3  600 

White  pine  

320 

480 

960 

Yellow  pine 

500 

730 

1  460 

Hemlock                 .  . 

240 

360 

720 

Oak  

400 

600 

1  200 

the  value  of  S.     For  a  solid  rectangular  beam,  Table  XI  gives  R  =  —^-. 

Substituting  the  width  and  depth  of  the  beam  in  this  last  formula,  we 

4  ~x  8'2 
get  R  =  -^—  =  42.67,    nearly.     The   length  of  the  beam   in   inches 

is  8  X  12  =  96  inches.  Substituting  all  the  values  in  the  formula  for 
the  safe  working  load,  we  get 

4  X  7,500  X  42.67 
W=—  —  —  13,334  lb.,  nearly.     Ans. 

EXAMPLE  2. — An  I  beam  having  a  depth  of  10  inches  and  an  area  of 
section  of  7.5  square  inches  is  to  be  used  as  a  cantilever  to  bear  a  vary- 
ing load  that  is  to  be  uniformly  distributed.  The  beam  is  made  of 
steel  and  is  100  inches  long.  What  safe  load  will  it  carry  ? 

SOLUTION.— From  Table  IX,  it  appears  that  the  proper  formula  to 

O     C    7"> 

be  used  is  W  —  — ^ — .  According  to  Table  X,  the  value  of  5  for  steel 
for  a  gradually  applied  load  is  7,500.  By  Table  XI  for  an  I  beam, 
R  =  5-55.  Substituting  values  in  this  last  formula,  we  get  7i  =  '  — 

o.oo  o.oo 

=  22.52,  nearly.  Substituting  values  in  the  formula  for  the  safe  work- 
ing load,  we  get 


2  X  7,500  X  22.52 
100 


=  3,378  lb.     Ans. 


TABLE  XI. 


VALUES  OF  R. 


Section. 


K. 


B 

T 


T 


b  h*  -  V  /i's 


AD 


AD1  -ad* 
8  D 

bl? 
24 


A  h 


A  h 


A  h 


A  h 


A  h 
3.33 


A  h 
3.67 


28  STRENGTH  OF  MATERIALS.  §  7 

46.  While  the  formulas  given  in  Table   IX  will  allow 
the    safe    load    on    a   given    beam    to    be    calculated,    they 
cannot   readily    be    transformed    to    give    directly    the    size 
of  beam  required  under  given  conditions  to  carry  a  given 
load. 

In  practice,  when  it  is  required  to  determine  the  size  of 
beam,  the  length  between  supports  is  known.  After  the 
shape  of  cross-section  of  the  beam  has  been  chosen,  dimen- 
sions for  the  beam  may  be  assumed  and  its  safe  load  for  the 
given  dimensions  calculated.  If  the  load  thus  found  falls 
below  the  load  the  beam  is  to  carry,  larger  dimensions  must 
be  chosen.  In  case  of  rolled  sections,  such  as  angle  irons, 
T  irons,  channels,  or  I  beams,  catalogues  of  manufacturers 
should  be  consulted  and  standard  sizes  chosen.  These  cata- 
logues usually  contain  the  area  of  the  section  for  different 
weights  and  dimensions  of  rolled  sections. 

47.  The  values  of  R  given  in  Table  XI  apply  only  when 
the  dimension  marked  JL  is  vertical.     If  the  beam  is  placed 
in  any  other  position,  R  will  have  a  different  value,  which 
can  only  be  calculated  by  the  aid  of  higher  mathematics. 
Beams  rigidly  fixed  at  both  ends  are   rarely  met   with  in 
practice,  and  it  is  safer  to  assume  that  the  beam  merely 
rests  on  two  supports. 


EXAMPLES  FOR  PRACTICE. 

1.  What  weight  will  a  yellow-pine  rectangular  beam  carry  safely 
under  a  steady  load  when  used  as  a  cantilever  and  uniformly  loaded  ? 
The  beam  is  9  feet  6  inches  long,  16  inches  deep,  and  4  inches  wide. 

Ans.  4,371  Ib. 

2.  What  weight  would  the  beam   in  example  1  carry  safely  if  it 
rested  on  two  supports  ?  Ans.  17,484  Ib. 

3.  What  weight  can  safely  be  carried  at  the  end  of  a  wrought-iron 
round  cantilever  3  inches  in  diameter  when  the  load  is  suddenly  applied 
4  feet  2  inches  from  the  support  of  the  cantilever  ?  Ans.  212  Ib. 

4.  What  steady  load  can  be  carried  safely  by  the  same  beam  given 
in  the  preceding  example  ?  Ans.  726  Ib. 


STRENGTH  OF  MATERIALS. 


29 


SHEARING,  OR  CUTTING,   STRENGTH 
OF  MATERIALS. 

48.  The  shearing  strength  of  any  material  is  the 
resistance  offered  by  its  fibers  to  being  cut  in  two.  Thus, 
the  pressure  of  the  cutting 
edges  of  an  ordinary  shear- 
ing machine,  Fig.  0,  causes 
a  shearing  stress  in  the 
plane  a  b.  The  unit  shear- 
ing force  may  be  found  by 
dividing  the  force  P  by  the 
area  of  the  plane  a  b. 

Fig.    7    shows   a    piece   in 
double  shear;   here  the  cen- 
tral piece  c  d  is   forced   out  FIG.  6. 
while  the  ends  remain  on  their  supports  M  and  N. 

The  shearing  strength  of  any  body  is  directly  proportional 
to  its  area. 


ILL 


49.  In  Table  XII    are  given  the  greatest  and  the  safe 
shearing   strengths    per   square    inch   of   different  kinds  of 
materials. 

50.  Rules  for  Shearing. — In  general,  the  force  required 
to  shear  a  piece  of  material  in  double  shear  will  be  twice  that 
required  for  single  shear.     This  does  not  apply  to  all  cases, 
however;    a  notable   exception   are  rivets  in  double  shear. 
Experiments  have  determined  that  the    force  required    to 
shear   both  iron  and  steel  rivets  in  double  shear  averages 
about  1.85  times  that  required  for  single  shear. 


30 


STRENGTH  OF  MATERIALS. 


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*  ^ 

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fe    t 

*  _d 

h  S 

1C    »O    »C    O    O 

£  t 
P  'J 

if 

4j  o 

<??  ^  <N  co  »o 

Average 

ci,,.. 

^s 

3  02 

Parallel 
to  Grain. 

oooooooooo 

oooooooooo 

OOOOOCO-^G<JCOO 

»o"  oo"  rjT  ^jT  <pT 

«    CO    rj(    ^*    10 



,      .§        : 

Materia 

•c  ^  I°i3    :    : 

c  S  o  «    :    : 

•     0   .>     ^    to       .     w 

•  .b   VH    ^  ^-,    <u   c 
d  -M  13   >,  £  -S  'a^ 

O    rC     a;     l-i     3     O-i           O 

t.   bo  •»-»   *  5   *T  te   5   ti 

•^  g  2  .s  !  j§  j  !  3 

O^c/2Oc^^!>*Wc/2O 

§  7  STRENGTH  OF  MATERIALS.  31 

Let       a  —  area  of  cross-section  in  square  inches; 
5  =  shearing  stress  as  given  in  table  ; 
IV  =  load  in  pounds. 

Then,  to  find  the  safe  load  or  the  load  that  will  shear  the 
material  : 

Rule  9.  —  Multiply  the  area  of  the  section  by  the  shearing 
stress. 

Or,  W=aS. 

In  applying  this  rule,  it  should  be  remembered  that  when 
the  safe  load  the  material  can  bear  is  required,  the  value 
of  S  is  to  be  taken  from  the  column  headed  "  Safe  Loads," 
selecting  the  safe  load  corresponding  to  the  nature  of  the 
load.  When  it  is  desired  to'  find  the  load  at  which  the  mate- 
rial will  fail,  the  value  of  5  is  to  be  taken  from  the  column 
headed  "Ultimate  Shearing  Strength." 

EXAMPLE.  —  If  the  beam  in  double  shear  shown  in  Fig.  7  is  rect- 
angular and  measures  4  in.  X  2  in.,  what  steady  safe  load  would  you 
allow  if  the  beam  were  made  of  structural  steel  ? 

SOLUTION.  —  According  to  the  table,  a  safe  load  of  10,000  pounds  per 
square  inch  of  section  may  be  allowed.  Applying  rule  9  and  multi- 
plying by  2,  since  the  beam  is  in  double  shear,  we  get 

W  =  4  X  2  X  10,000  X  2  =  160,000  Ib.     Ans. 

51.  To  find  the  area  required  for  a  material  subjected 
to  a  shearing  stress: 

Rule  1O.  —  Divide  the  load  by  the  shearing  stress. 


In  applying  rule  1O,  it  should  be  remembered  that  for 
double  shear  the  result  is  to  be  divided  by  2  to  obtain  the 
area  of  the  beam. 

EXAMPLE.  —  A  white  pine  beam  is  subjected  to  a  suddenly  applied 
shearing  stress  by  a  load  of  4,000  pounds,  the  beam  being  in  double 
shear  and  the  load  being  applied  across  the  grain.  What  should  be 
the  area  of  the  beam  to  bear  this  stress  safely  ? 


32  STRENGTH  OF  MATERIALS.  §  7 

SOLUTION.— By  the  table,  S=  250.     Then,  by  rule  1O,  we  get 

a  =    '         =  16  square  inches  for  single  shear. 
250 

Dividing  the  result  by  2  for  double  shear,  we  get  16  -i-  2  =  8  sq.  in. 
area.     Ans. 


EXAMPLES    FOR    PRACTICE. 

1.  A  hemlock  beam  8  in.  X  2  in.  is  in  single  shear  parallel  to  the 
grain.     What  load  will  cause  it  to  fail  by  shearing  ?  Ans.  4,000  Ib. 

2.  A  wrought-iron  rivet  1  inch  in  diameter  is  in  double  shear.     At 
what  load  will  it  be  likely  to  shear  ?  Ans.  55,214  Ib.,  nearly. 


TORSIOK. 

52.  When  a  force  is  applied  to  a  beam,  bar,  or  rod  in 
such  a  manner  that  it  tends  to  twist  it,  the  stress  thus  pro- 
duced is  termed  torsion.  Torsion  manifests  itself  in  the 
case  of  rotating  shafts,  such  as  line  shafts  and  engine  shafts. 


LINE  SHAFTING. 

53.  A  line  of  shafting  is  one  continuous  run,  or  length, 
composed  of  lengths  of  shafts  joined  together  by  couplings. 

54.  The  main  line  of  shafting  is  that  which   receives 
the  power  from  the  engine  or  motor  and  distributes  it  to 
the  other  lines  of  shafting  or  to  the  various  machines  to  be 
driven. 

Line  shafting  is  supported  by  hangers,  which  are  brackets 
provided  with  bearings,  bolted  either  to  the  walls,  posts, 
ceilings,  or  floors  of  the  building.  Short  lengths  of  shafting, 
called  countershafts,  are  provided  to  effect  changes  of 
speed  and  to  enable  the  machinery  to  be  stopped  or  started. 

55.  Shafting  is  usually  made   cylindrically  true,  either 
by  a  special  rolling  process,  when  it  is  known  as  cold-rolled 
shafting,  or  else  it  is  turned  up  in  a  machine  called  a  lathe; 
In  the  latter  case  it  is  called  bright  shafting.     What  is 


STRENGTH  OF  MATERIALS. 


33 


known  as  black  shafting  is  simply  bar  iron  rolled  by  the 
ordinary  process  and  turned  where  it  receives  the  coup- 
lings, pulleys,  bearings,  etc. 

56.  The    diameter  of  bright  turned   shafting  increases 
by  £  inch   up  to  about   3£  inches  in  diameter;  above  this 
diameter  it  increases  by  £  inch.     The  actual  diameter  of  a 
bright  shaft  is  -fa  inch  less  than  the  commercial  diameter,  it 
being  designated  from  the  diameter  of  the  ordinary  round 
bar  iron  from  which  it  is  turned.     Thus,  a  length  of  what 
is  called  3-inch  bright  shafting  is  really  only  2f|  inches  in 
diameter. 

Cold-rolled  shafting  is  designated  by  its  commercial  diam- 
eter; thus,  a  length  of  what  is  called  3-inch  shafting  is 
3  inches  in  diameter. 

57.  In  the  following  table  is  given  the   maximum  dis- 
tance between  the  bearings  of  some  continuous  shafts  that 
are  used  for  the  transmission  of  power: 

TABLE  XIII. 


DISTANCE  BETWEEN  BEARINGS. 


Distance  Between  Bearings. 

Diameter 
of  Shaft. 
Inches. 

Feet. 

Wrought-Iron 
Shaft. 

Steel  Shaft. 

2 

11 

11.50 

3 

13 

13.75 

4 

15 

15.75 

5 

17 

18.25 

6 

19 

20.00 

7 

21 

22.25 

8 

23 

24.00 

9 

25 

26.00 

11.     X.     I.—2J 


34  STRENGTH  OF  MATERIALS.  §  7 

Pulleys  from  which  considerable  power  is  to  be  taken 
should  always  be  placed  as  close  to  a  bearing  as  possible. 

58.  The   diameters   of    the    different   lengths   of   shafts 
composing  a  line  of  shafting  may  be  proportional  to  the 
quantity  of  power  delivered   by  each  respective  length.      In 
this  connection  it  is  to  be  observed  that  the  positions  of  the 
various  pulleys  in  reference  to  the  bearings  must  be  taken 
into  consideration    in    deciding   upon    the    size    of  a    shaft. 
Suppose,   for  example,   that  a  piece  of  shafting  delivers  a 
certain  amount  of  power;  then,  it  is  obvious  that  the  shaft 
will    deflect   or   bend    less  if   the   pulley  transmitting    that 
poAver  be  placed  close  to  a  hanger  or  bearing  than  if  it  be 
placed  midway  between  the  two  hangers  or  bearings.     It  is 
impossible  to  give  any  rule  for  the  proper  distance  of  bear- 
ings that  could  be  used  universally,  as  in  some  cases  the 
requirements  demand  that  the  bearings  be  nearer  than  in 
others. 

Wherever  possible  it  is  advisable  to  have  the  main  line  of 
shafting  run  through  the  center  of  the  room,  or  at  least  far 
enough  from  either  wall  to  allow  countershafts  to  be  placed 
on  either  side  of  it.  When  this  is  done,  power  may  be  taken  off 
the  main  shaft  from  either  side  by  alternate  pulleys,  and  the 
deflection  caused  in  the  main  shaft  in  one  direction  by  one 
pulley  will  be  counteracted  by  the  deflection  caused  in  the 
opposite  direction  by  the  next  pulley. 

If  the  work  done  by  a  line  of  shafting  is  distributed  quite 
equally  along  its  entire  length  and  the  power  can  be  applied 
near  the  middle,  the  strength  of  the  shaft  need  be  only  half 
as  great  as  would  be  required  if  the  power  were  applied  at 
one  end. 

59.  To  compute  the  horsepower  that  can  be  transmitted 
by  a  shaft  of  any  given  diameter  : 

Let  D  =  diameter  of  shaft ; 

R  —  revolutions  per  minute; 

H '=  horsepower  transmitted; 

C  —  constant  given  in  Table  XIV. 


§  7  STRENGTH  OF  MATERIALS. 

TABLE  XIV. 


3S 


CONSTANTS  FOR  LINE  SHAFTING. 


Material  of  Shaft. 

No  Pulleys 
Between 

Pulleys 
Between 

Bearings. 

Bearings. 

Steel  or  cold-rolled  iron 

65 

85 

Wrought  iron  

70 

95 

Cast  iron  

90 

120 

In  the  above  table  the  bearings  are  supposed  to  be  spaced 
so  as  to  relieve  the  shaft  of  excessive  bending;  also,  in  the 
third  vertical  column,  an  average  number  and  weight  of 
pulleys  and  power  given  off  is  assumed. 

6O.  Rules  and  Formulas. — In  determining  the  above 
constants,  allowance  has  been  made  to  insure  the  stiffness  as 
well  as  strength  of  the  shaft.  Cold-rolled  iron  is  consider- 
ably stronger  than  ordinary  turned  wrought  iron;  the 
increased  strength  is  due  to  the  process  of  rolling,  which 
seems  to  compress  the  metal  and  so  make  it  denser,  not 
merely  skin  deep,  but  practically  throughout  the  whole 
diameter.  We  have,  then,  the  following 

Rule  11. —  The  horsepower  that  a  shaft  will  transmit  equals 
the  product  of  the  cube  of  the  diameter  and  the  number  of 
revolutions,  divided  by  the  value  of  C  for  the  given  material. 


Or, 


D* 


c 


EXAMPLE.— What  horsepower  will  a  3-inch  wrought-iron  shaft  trans- 
mit which  makes  100  revolutions  per  minute,  there  being  no  pulleys 
between  bearings  ? 

SOLUTION. — Applying  rule  11  and  substituting,  we  have 


3  X  3  X  3  X  100 
70 


=  38.57  H.  P.     Ans. 


36  STRENGTH  OF  MATERIALS.  §  7 

If  there  were  the  usual  amount  of  power  taken  off,  as  mentioned 
above,  we  should  take  C  =  95. 


Then,  H  =  ::^±—  =  28.42  H.  P.     Ans. 

61.  To  compute  the  number  of  revolutions  a  shaft  must 
make  to  transmit  a  given  horsepower : 

Rule  12. —  The  number  of  revolutions  necessary  for  a  given 
horsepoiver  equals  the  product  of  the  value  of  C  for  the  given 
material  and  the  number  of  horsepower,  divided  by  the  cube 
of  the  diameter. 

Or,  *  =  %£ 

EXAMPLE. — How  many  revolutions  must  a  3-inch  wrought-iron 
shaft  make  per  minute  to  transmit  28.42  horsepower,  power  being 
taken  off  at  intervals  between  the  bearings  ? 

SOLUTION. — Applying  the  rule  just  given  and  substituting,  we  have 


62.  To  compute  the  diameter  of  a  shaft  that  will  trans- 
mit a  given  horsepower,  the  number  of  revolutions  the  shaft 
makes  per  minute  being  given: 

Rule  13.  —  The  diameter  of  a  shaft  equals  t/ie  cube  root 
of  the  quotient  obtained  by  dividing  the  product  of  the  value 
of  C  for  the  given  material  and  the  number  of  horsepower  by 
the  number  of  revolutions. 

Or,  D  = 

EXAMPLE.  —  What  must  be  the  diameter  of  a  wrought-iron  shaft  to 
transmit  38.57  horsepower,  the  shaft  to  make  100  revolutions  per 
minute,  no  power  being  taken  off  between  bearings  ? 

SOLUTION.  —  By  rule  13,  we  have 

x 


D  =  j  '      =  ^27  =  3  in.     Ans. 

63.  As  the  speed  of  shafting  is  used  as  a  multiplier  in 
the  calculations  of  the  horsepower  of  shafts,  it  is  readily 
seen  that  a  shaft  having  a  given  diameter  will  transmit  more 


§  7  STRENGTH  OF  MATERIALS.  37 

power  in  proportion  as  its  speed  is  increased.  Thus,  a  shaft 
that  is  capable  of  transmitting  10  horsepower  when  making 
100  revolutions  per  minute  will  transmit  20  horsepower  when 
making  200  revolutions  per  minute.  We  may,  therefore,  say 
the  number  of  horsepower  transmitted  by  a  shaft  is  directly 
proportional  to  the  number  of  revolutions. 


EXAMPLES    FOB    PRACTICE. 

1.  What   horsepower  will  a  2^-inch  wrought-iron  shaft  transmit 
when  running  at  110  revolutions  per  minute,  it  being  used  for  trans- 
mission only  ?  Ans.  24.55  H.  P. 

2.  A  6-inch  cast-iron  shaft  transmits  150  horsepower.     How  many 
revolutions  per  minute  must  it  make,  no  power  being  taken  off  between 
bearings  ?  Ans.  62£  R.  P.  M. 

3.  What  should  be  the  diameter  of  a  wrought-iron  shaft  to  transmit 
100  horsepower  at  150  revolutions  per  minute,  power  being  taken  off 
between  bearings  ?  Ans.  4  in.,  nearly. 

4.  The  machines  driven  by  a  certain  line  of  wrought-iron  shafting 
take  their  power  from  various  points  between  the  bearings,  and  if  all 
were   working  together  at  their  full   capacity,    they   would  require 
65  horsepower  to  drive  them.     What  diameter  should  the  shaft  be  if  it 
runs  at  150  revolutions  per  minute  ?  Ans.  3£  in.,  nearly. 


A  SERIES 

OF 

QUESTIONS   AND   EXAMPLES 

RELATING   TO  THE    SUBJECTS 
TREATED  OF  IN  THIS  VOLUME. 


It  will  be  noticed  that  the  Examination  Questions  that 
follow  have  been  divided  into  sections,  which  have  been 
given  the  same  numbers  as  the  Instruction  Papers  to  which 
they  refer.  No  attempt  should  be  made  to  answer  any  of 
the  questions  or  to  solve  any  of  the  examples  until  that 
portion  of  the  text  having  the  same  section  number  as  the 
section  in  which  the  questions  or  examples  occur  has  been 
carefully  studied. 


ARITHMETIC. 

(PART  1.) 


EXAMINATION  QUESTIONS. 

(1)  What  is  arithmetic  ? 

(2)  What  is  a  number  ? 

(3)  What  is  the  difference  between  a  concrete  number 
and  an  abstract  number  ? 

(4)  Define  notation  and  numeration. 

(5)  Write  each  of  the  following  numbers  in  words:     980; 
605;  28,284;  9,006,042;  850,317,002;  700,004. 

(6)  Represent  in  figures  the  following  expressions: 
Seven  thousand  six  hundred.      Eighty-one  thousand  four 

hundred  two.  Five  million  four  thousand  seven.  One 
hundred  eight  million  ten  thousand  one.  Eighteen  million 
six.  Thirty  thousand  ten. 

(7)  What  is  the  sum  of    3,290,   504,   865,403,  2,074,  81, 
and  7?  Ans.  871,359. 

(8)  709  +    8,304,725  +    391    +    100,302  +  300  +  909 
=  what?  Ans.  8,407,336. 

(9)  During  a  12-hour  test  of  a  steam  engine  the  counter 
showed  the  number  of  revolutions  per  hour  to  have  been  as 
follows:     12,600,    12,444,    12,467,     12,528,    12,468,    12,590, 
12,610,    12,589,   12,576,   12,558,  12,546,  12,532.      How  many 
revolutions  were  made  during  the  test  ?      Ans.   150,508  rev. 

§1 

For  notice  of  copyright,  see  page  immediately  following  the  title  page. 


2  ARITHMETIC.  §  1 

(10)  Find    the  difference    between    the   following:     (a) 
50,962  and  3,338;  (b)  10,001  and  15,339. 

.         (  (a)     47,624 
3'  (  (b)     5,338. 

(11)  (a)  70,968  -  32,975  =  ?   (b)  100,000  -  98,735  =  ? 

((*)  37,993. 
M(*)  1,265. 

(12)  On  a  certain  morning  7,240   gallons  of  water  were 
drawn  from  an  engine-room  tank  and  4,780   gallons  were 
pumped  in.     In  the  afternoon  7,633  gallons  were  drawn  out 
and  8,675  gallons  pumped  in.      How  many  gallons  remained 
in  the  tank  at  night,  if  it  contained  3,040  gallons  at  the  begin- 
ning of  the  day  ?  Ans.   1,622  gal. 

(13)  Find  the  product  of  the  following :     (a)  526,387  X  7; 
(b)  700,298  X  17;  (c)  217  X  103  X  67. 

f  (a)     3,684,709. 
Ans.  ]  (b)      11,905,066. 
(  (c)      1,497,517. 

(14)  If  your  watch  ticks  once  every  second,  how  many 
times  will  it  tick  in  one  week  ?  Ans.   604,800. 

(15)  An  engine  and  boiler  in  a  manufactory  are  worth 
$3,246.     The  building  is  worth  three  times  as  much,  plus 
$1,200,  and  the  tools  are  worth  twice  as  much  as  the  build- 
ing, plus  $1,875.      (a)  What  is  the  value  of  the  building  and 
tools  ?     (b}  What  is  the  value  of  the  whole  plant  ? 

I  (a)     $34,689. 
Sl((£)      $37,935. 

(16)  Divide  the  following : 

(a)  962,842  by   84;   (b)  39,728  by  63;  (c)  29,714  by  108; 
(d)  406,089  by  135.  f  (a)     11,462.4047. 

J   (b}     630.603. 
AnS>  *   (c)     275.1296. 
(</)    3,008.0666. 

(17)  Suppose    that   in    1    hour    10    pounds   of    coal    are 
burned  per  square  foot  of  grate  area  in  a  certain  boiler, 


§  1  ARITHMETIC.  3 

and  that  9  pounds  of  water  are  evaporated  per  pound  of 
coal  burned.  If  the  grate  area  is  30  square  feet,  how  many 
pounds  of  water  would  be  evaporated  in  a  day  of  10  hours  ? 

Ans.   27,000  Ib. 

(18)  If  a  mechanic  receives  $1,500  a  year  for  his  labor 
and   his  expenses  are  $968   per  year,  in  what  time  can  he 
save  enough  to  buy  28  acres  of  land  at  $133  an  acre  ? 

Ans.  7  years. 

(19)  Solve  the  following  by  cancelation : 
(a)   (72  X  48  X28  X  5)  -5-  (84  X  15  X  7  X  6). 

(If)   (80  X  60  X  50  X  16  X  14)  -*•  (70  X  50  X  24  X  20). 

(«)     9f 
W     88. 

(20)  A   freight    train    ran    365    miles  in  one  week,   and 
3  times  as  far,  lacking  246  miles,  the  next  week;  how  far 
did  it  run  the  second  week  ?  Ans.   849  miles. 

(21)  If  the  driving  wheel  of  a  locomotive  is  16  feet  in 
circumference,  how  many  revolutions  will  it  make  in  going 
from  Philadelphia  to  Pittsburg,  the  distance  between  them 
being  354  miles,  and  there  being  5,280  feet  in  1  mile  ? 

Ans.   116,820  rev. 


ARITHMETIC. 

(PART  2.) 


EXAMINATION    QUESTIONS. 

(1)  What  is  a  fraction? 

(2)  What  are  the  terms  of  a  fraction  ? 

(3)  What  does  the  denominator  show  ? 

(4)  What  does  the  numerator  show  ? 

(5)  Is  Y-  a  proper  or  an  improper  fraction,  and  why  ? 

(6)  Write  three  mixed  numbers. 

(7)  Reduce  the  following  fractions  to  their  lowest  terms: 

I,  TV  A,  If-  Ans-  i»  i.  i.  *• 

(8)  Reduce   6   to   an  improper  fraction  whose   denomi- 
nator is  4.  Ans.  -Sjk. 

(9)  Reduce  7|,  13^,  and  lOf  to  improper  fractions. 

Ans.  V,  W,  ¥• 

(10)  How  many  feet  does  the  piston  of  a  steam  engine 
pass  over  in  a  week  of  6  days,  running  8£  hours  a  day,  if 
the  length  of  the  stroke  of  the  engine  is  1|-  feet  and  the 
number  of  revolutions  per  minute  is  160  ? 

Ans.  1,468,800  ft. 

(11)  i  +  l  +  l=?  Ans.  1. 

?  Ans.       - 


(13)     42  +  31|  +  9TV  =  ?  Ans. 

§1 

For  notice  of  copyright,  see  page  immediately  following  the  title  page. 


ARITHMETIC. 


§1 


(15) 


(14)  An  iron  plate  is  divided  into  four  sections ;  the  first 
contains  29|  square  inches;  the  second,  50f  square  inches; 
the  third,  41  square  inches;  and  the  fourth,  G9T37  square 
inches.  How  many  square  inches  are  in  the  plate  ? 

Ans.   190TV  sq.  in. 

What  is  the  difference  between  $-  and  T\  ?  13  and  7TV  ? 
229^-?  Ans.  yV;  5^;  83||. 

(16)  Solve  the  following : 

(a)   35  -f-  fa     (b]   yV-3;     (c)    y  -  9;     (</)    ^  -i- fa 

(a)      112. 

(*)        TV 

Ans.        (c]      ff. 

(')      3|.8' 

(17)  The  numerator  of  a  fraction  is  28  and  the  value  of 
the  fraction  is  ^ ;  what  is  the  denominator  ?  Ans.  32. 

(18)  Four  bolts  are  required,  2£,  6^,  3TV,  and  4  inches 
long.     How  long  a  piece  of  iron  will  be  required  from  which 
to  cut  them,  allowing  ^  of  an  inch  to  each  bolt  for  cutting 
off  and  finishing  ?  Ans.  ISy3^  in. 


ARITHMETIC. 

(PART  3.) 


EXAMINATION  QUESTIONS. 

(1)  State  the  difference  between  a  common  fraction  and 
a  decimal  fraction. 

(2)  Reduce  the  following  fractions  to  equivalent  decimals : 

i>  &>  T2»   TlHSy  and  TTSFO-  -5. 

.875. 

Ans.      .15625. 
,65. 
,125. 

(3)  Write  out  in  words  the  following  numbers:  .08,  .131, 
.0001,  .000027,  .0108,  and  93.0101. 

(4)  What   is  the   sum  of  .125,    .7,    .089,    .4005,   .9,    and 
.000027  ?  Ans.  2.214527. 

(5)  Add  17  thousandths,  2  tenths,  and  47  millionths. 

Ans.   .217047. 

(6)  Work  out  the  following  examples: 

(a)  709.63  — .8514;  (b}  81.963  —  1.7;  (c]  18  — .18;  (d] 
1  -  .001 ;  (e)  872.1  -  (.8721  +  .008) ;  (/)  (5.028  +  .0073) 
-  (6.704  -  2.38).  {  (a)  708.7786. 

(b}     80.263. 

(c)  17.82. 

(d)  .999. 

(e)  871.2199. 
,  (/)  .7113. 

§1 

For  notice  of  copyright,  see  page  immediately  following  the  title  page. 


Ans.  -< 


8  ARITHMETIC.  §  1 

(7)  The  cost  of  the  coal  consumed  under  a  nest  of  steam 
boilers  during  a  week's  run  was  as  follows:  Monday,  $15.83; 
Tuesday,    $14.70;    Wednesday,   $14.28;    Thursday,   $13.87; 
Friday,   $14.98;    Saturday,  $12.65.     What  was  the  cost  of 
the  week's  supply  of  coal  ?  Ans.  $86.31. 

(8)  It  is  desired  to  increase  the  capacity  of  an  electric- 
light  plant  to  1,500  horsepower  by  adding  a  new  engine.     If 
the  indicated   horsepower   of   the    engines    already    in    use 
is  482£,  316£,  and  390f,  what  power  must  the  new  engine 
develop?  Ans.   310.11$  H.  P. 

(9)  The  inside  diameter  of  a  6-inch  steam  pipe  is  6.06 
inches,  and  the  outside  diameter  is  6.62  inches.      How  thick 
is  the  pipe  ?  Ans.  .28  in. 

(10)  Find   the   products   of   the    following   expressions: 
(a)    .  013  X.  107;    (b)    203  X  2.03  X  .203;     (c)    (2.7x31.85) 
X  (3.  16  -.316);  (d)  (107.8  +  6.541-31.96)  X  1.742. 

(a)  .001391. 

,    (b]  83.65427. 

1    (c)  244.56978. 

(d)  143.507702. 

(11)  How  many  square  feet  of  heating  surface  are  in  the 
tubes  of  a  boiler  having  sixty  3-inch  tubes,  each  15£  feet 
long,  if  the  heating  surface  of  each  tube  per  foot  in  length  is 
.728  square  foot  ?  Ans.   677.04  sq.  ft. 

(12)  Find  the  values  of  the  following  expressions  : 

7       ...    U     ,  N    1.25  X  20  X  3  (  (a)     37$. 

(«)    S;    (*)   f  ;    <<)         87  +  88       •        Ans.       (q     .75 


459  +  32  (c)      210f 

(13)  The   distance  around  a  cylindrical  boiler  is  166.85 
inches.     If  there  are  72  rivets  in  one  of  the  circular  seams, 
find  what  the  pitch  (distance  between  the  centers  of  any  two 
rivets)  of  the  rivets  is.  Ans.  2.317+  in. 

(14)  A  keg  of  £*  x  21"  boiler  rivets  weighs  100  pounds 
and  contains  133  rivets.     What  is  the  weight  of  each  rivet  ? 

Ans.  .75+lb. 


§  1  ARITHMETIC.  9 

(15)  How  many  inches  in  .875  of  a  foot  ?        Ans.  10£  in. 

(16)  What  decimal  part  of  a  foot  is  T3-g-  inch  ? 

Ans.    .015625  ft. 

(17)  In  a  steam-engine  test  of  an  hour's  duration,  the 
horsepower  developed  was  found  to  be  as  follows  at  10-min- 
ute  intervals:  48.63,   45.7,  46.32,   47.9,  48.74,   48.38,  48.59. 
What  was  the  average  horsepower  ?          Ans.  47.75+  H.  P. 


s.    I.—2J, 


ARITHMETIC. 

(PART  4.) 


EXAMINATION  QUESTIONS. 

(1)  What  is  25$  of  8,428  Ib.?  Ans.   2,107  Ib. 

(2)  What  is  |$  of  $35,000?  Ans.  $175. 

(3)  What  per  cent,  of  50  is  2  ?  Ans.   4$. 

(4)  What  per  cent,  of  10  is  10  ?  Ans.   100$. 

(5)  The  coal  consumption  of  a  steam  plant  is  5,500  Ib.  per 
day  when  the  condenser  is  not  running,  or  an  increase  of 
15$  over  the  consumption  when  the  condenser  is  used.     How 
many  pounds  are  used  per  day  when  the  condenser  is  run- 
ning ?  Ans.  4,782.61  Ib. 

(6)  An  engineer  receives  a  salary  of  $950.     He  pays  24$ 
of  it  for  board,  12£$    of  it  for  clothing,   and  17$  of  it  for 
other  expenses.     How  much  of  it  does  he  save  a  year  ? 

Ans.  $441.75. 

(7)  If  37£$  of  a  number  is  961.38,  what  is  the  number  ? 

Ans.   2,563.68. 

(8)  The  speed  of  an  engine  running  unloaded  was  1£$ 
greater  than  when  running  loaded.     If  it  made  298  revolu- 
tions per  minute  with  the  load,  what  was  its  speed  running 
unloaded  ?  Ans.   302.47  rev.  per  min. 

(9)  Reduce  4  yd.  2  ft.  10  in.  to  inches.  Ans.   178  in. 

§2 

For  notice  of  copyright,  see  page  immediately  following  the  title  page. 


2  ARITHMETIC.  §  2 

(10)  Reduce  3,722  in.  to  higher  denominations. 

Ans.   103  yd.  1  ft.  2  in. 

(11)  Reduce  764,325  cu.  in.  to  cubic  yards. 

Ans.   16  cu.  yd.   10  cu.  ft.   549  cu.  in. 

(12)  A  carload  of  coal  weighed  16  T.  8  cwt.  75  Ib.      How 
many  pounds  did  this  amount  to  ?  Ans.   32,875  Ib. 

(13)  Reduce  25,396  Ib.  to  higher  denominations. 

Ans.  12  T.  13  cwt.  96  Ib. 

(14)  What  is  the  sum  of  2  yd.  2  ft.  3  in.,  4  yd.  1  ft.  9  in., 
and  2  ft.  7  in.  ?  Ans.  8  yd.  7  in. 

(15)  From  a  barrel  of  machine  oil  is  sold  at  one  time 
10  gal.  2  qt.  1  pt.  and  at  another  time  16  gal.  3  qt.     How 
much  remained  ?  Ans.  4  gal.  1  pt. 

(16)  If  1   iron  rail  is  17  ft.   3  in.   long,  how  long  would 
51  such  rails  be  if  placed  end  to  end  ?          Ans.  879  ft.  9  in. 

(17)  Multiply  3  qt.  1  pt.  by  4.7.  Ans.   32.9  pt. 

(18)  A  main  line  shaft  is  composed  of  four  lengths  each 
15ft.  Sin.  long;  one  length  14ft.  Sin.  long;  and  one  length 
8  ft.  10  in.  long.     There  are  six  hangers  spaced  equally  dis- 
tant apart,  one  being  placed  at  each  extremity  of  the  shaft, 
8  in.  from  the  ends.     What  is  the  distance   between  the 
hangers  ?  Ans.   16  ft.  9  j  in. 

(19)  If  the  length  of  a  boiler  shell  is  18  ft.  11£  in.,  how 
many    rivets   should  there   be  in  one   of   the   longitudinal 
seams  if  it  is  a  single-riveted  seam,  supposing  the  rivets  to 
be  1^  in.  between  centers  and  the  two  end  rivets  to  be  1£  in. 
from  each  end  of  the  boiler  ?  Ans.   181  rivets. 


ARITHMETIC. 

(PART  5.) 


EXAMINATION  QUESTIONS. 

(1)  What  is  the  square  of  108  ?  Ans.   11,664. 

(2)  What  is  the  cube  of  181.25? 

Ans.  5,954,345.703125. 

(3)  Find  the  values   of   the   following:    (a)  .01333;    (b) 
301.011s;  (0   (*)•;   (d)   (3f)°. 

' (a)      .000002352637. 
(b)      27,273,890.942264331. 

O    A. 

(d)     52|i,  or  52.734375. 

NO'TE.— In  the  answers  to  the  following  examples  requiring  the 
extraction  of  a  root,  a  minus  sign  after  a  number  indicates  that  the  last 
digit  is  not  quite  as  large  as  the  number  printed.  Thus,  12,497—  indi- 
cates that  the  number  really  is  12.496  +  ,  and  that  the  6  has  been  made 
a  7  because  the  next  succeeding  figure  was  5  or  greater.  The  answers 
to  the  examples  requiring  the  extraction  of  cube  root  are  exact  to  the 
last  digit  given.  If  the  student  is  unable  to  get  the  same  answer  by 
means  of  the  method  described  in  Arts.  3O-34,  he  should  then  use 
the  method  described  in  Arts.  35  and  36. 

(4)  Find  the  square  root  of  the  following:    (a)  3,486,^ 
784.401;  (b)  9,000,099.4009.  I  (a)     1,867.29+. 

3'  1  (b)     3,000.017-. 

(5)  Extract  the  cube  root  of  .32768.  Ans.   .6894+. 

(6)  Extract  the  cube  root  of  2  to  four  decimal  places. 

Ans.   1.2599+. 

§2 

For  notice  of  copyright,  see  page  immediately  following  the  title  page 


ARITHMETIC. 


(7)  Solve  the  following:   (a)  f/123.21;  (b) 

Ans. 

(8)  Solve  the  following :  (a) 


1/114.921. 

(a)  11.1. 

(b)  10.72+. 


'.0065;  (b)  |/.  000021. 

j  («)      .18663-. 
3'  (  (J)      .02759-. 
Find  the  value  of  ;r  in  the  following : 

(9)  11.7  :  13  ::  20  :  x.  Ans.  22.22+. 

(10)  (a)  20  +  7  :  10+  8  ::  3  :  x\  (b}  12a  :  100'  ::  4  :  x. 

MS  " 


(ii) 


4r      21'  ™'  24      16'   ^  10      100' 


277.7+. 
60 


150 


_£- 
600' 


Ans. 


15 
;45 
(a) 

w 

(rf) 

M 


12. 

12. 

20. 

180. 

40. 


(12)  If   a   piece   of   2-inch    shafting  3£  ft.    long  weighs 
37.45  lb.,  how  much  would  a  piece  6f  ft.  long  weigh  ? 

Ans.  72.225  lb. 

(13)  If  a  railway  train  runs  444  miles  in  8  hr.  40  min.,  in 
what  time  can  it  run  1,060  miles  at  the  same  rate  of  speed  ? 

Ans.   20  hr.  41.44  min. 

(14)  If  a  pump  discharging  135  gal.  per  min.  fills  a  tank 
in   38  min.,   how  long  would   it  take  a  pump   discharging 
85  gal.  per  min.  to  fill  it  ?  Ans.   60T6T  min. 

(15)  The  distances  around  the  driving  wheels  of  two  loco- 
motives are  12.56  ft.  and  15.7  ft.,  respectively.      How  many 
times  will  the  larger  turn  while  the  smaller  turns  520  times  ? 

Ans.  416  times. 

(16)  If  a  cistern  28  ft.  long,  12  ft.  wide,  10  ft.  deep  holds 
798  bbl.  of  water,  how  many  barrels  of  water  will  a  cistern 
hold  that  is  20  ft.  long,  17  ft.  wide,  and  6  ft.  deep  ? 

Ans.  484£  bbl. 


MENSURATION  AND  USE  OF 
LETTERS  IN  FORMULAS. 


EXAMINATION  QUESTIONS. 

A  =  5  h  —  200 

£  =  10  x  =  12 

i  =  3.5         Z>  =  120 

Work  out  the  solutions  to  the  following  formulas,  using 
the  above  values  for  the  letters: 

(1)     C=^~  An,   C=8. 

Q,44*£+J*  An,    6  = 


Ans.  v  =  4. 05+. 


n  <r=^    j.  ;, ,  ,;    -•        A».y- 


(5)  If   one  of  the   angles   formed    by  one  straight   line 
meeting  another  straight   line  equals  152°  3',   what  is  the 
other  angle  equal  to  ?  Ans.   27°  57'. 

(6)  Draw  an  obtuse  angle,  a  right  angle,  and  an  acute 
angle.     State  the  name  of  each  angle  by  using  letters  to 
designate  them. 

§3 

For  notice  of  copyright,  see  page  immediately  following  the  title  page 


2  MENSURATION  AND  §  3 

(7)  Draw  a  rhombus  and  then  draw  a  rectangle  having 
the  same  area. 

(8)  A  sheet  of   zinc  measures    11^    inches   by  2£  feet. 
How  many  square  inches  does  it  contain  ?     Ans.   345  sq.  in. 

(9)  How   many  boards  16   feet  long  and   5  inches  wide 
would  be  required  to  lay  a  floor  measuring  15  ft.  X  24  ft.  ? 

Ans.  54  boards. 

(10)  The  accompanying  figure  shows  the  floor  plan  of 
an  electric-light  station.     From  the  dimensions  given,  cal- 
culate the  number  of  square  feet  of  unoccupied  floor  space. 

Ans.   2,059.08  sq.  ft. 


Feed  Pump 

~~ 


Dynamos. 

>o 


Switchboard. 


(11)  A  triangle  has  three  equal  angles;  what  is  it  called  ? 

(12)  If  a  triangle  has  two  equal  angles,  what  kind  of  a 
triangle  is  it  ? 

(13)  In  a  triangle  ABC,  angle  ^4  =  23°  and  ,#  =  32°  32'; 
what  does  angle  C  equal  ?  Ans.   C  =  124°  28'. 


§3 


USE  OF  LETTERS  IN  FORMULAS. 


(14)  In  the  figure,  if  AD=  10  inches,  A  B  =  24  inches, 
and  B  C  =  13£  inches,  how  long  is  JD  £,  D  E  being  parallel 
to£C?  Ans.  D£=5.Q25  in. 

(15)  An  engine  room  is  52  feet  long 
and  39  feet  wide.      How  many  feet  is  it 
from  one  corner  to  a  diagonally  oppo- 
site one,  measured  in  a  straight  line  ? 

Ans.   65  ft. 

(10)  It  is  required  to  make  a  miter- 
box  in  which  to  cut  molding  to  fit 
around  an  octagon  post.  At  what  FIG.  n. 

angle  with  the  side  of  the  box  should  the  saw  run  ? 

Ans.   67£°. 

(17)  If  the  distance  between  two  opposite  corners  of  a 
hexagonal  nut  is  2  inches,  what  is  the  distance  between  two 
opposite  sides  ?  Ans.   1.732+ in. 

(18)  In  the  accompanying  figure,  if  the  distance  B I  is 

6  inches  and  H  K  18  inches,   what  is 
the  diameter  of  the  circle  ? 

Ans.   19.5  in. 

(19)     How   many   revolutions  will  a 
72-inch     locomotive     driver     make     in 
going  1  mile  ? 
FIG.  in.  Ans.  280.112  revolutions. 

(20)  A    pipe    has   an   internal    diameter  of   6.06  inches; 
what  is  the  area  of  a  circle  having  this  diameter  ? 

Ans.   28.8427  sq.  in. 

(21)  How  long  must  the  arc  of  a  circle  be  to  contain  12°, 
supposing  the  radius  of  the  circle  to  be  6  inches  ? 

Ans.   1.25664  in. 

(22)  What  is  the  area  of  the  sector  of  a  circle  15  inches 
in  diameter,  the  angle  between  the  two  radii  forming  the 
sector  being  12£°  ?  Ans.   6.1359  sq.  in. 

(23)  (a)  What  would  be  the  length  of  the  side  of  a  square 
metal    plate   having   an   area   of   103.8691    square   inches? 


4  MENSURATION  AND  §  3 

(b)  What  would  be  the  diameter  of  a  round  plate  having 
this  area  ?  (c]  How  much  shorter  is  the  circumference  of 
the  round  plate  than  Uie  perimeter  of  the  square  plate  ? 

(a)     10.1916  in. 
Ans.  •]  (£)      11 J  in. 
(c)      4. 638  in. 

(24)  Find  the  area  in  square  feet  of  the  entire  surface  of 
a  hexagonal  column  12  feet  long,  each  edge  of  the  ends  of 
the  column  being  4  inches  long.  Ans.   24.5774  sq.  ft. 

(25)  Find  the  cubical  contents  of  the  above  column  in 
cubic  inches.  Ans.   5,985.9648  cu.  in. 

(26)  Compute  the  weight  per  foot  of  an  iron  boiler  tube 
4  inches  outside  diameter  and  3.73  inches  inside  diameter, 
the  weight  of  the  iron  being  taken  at  .28  pound  per  cubic 
inch.  Ans.  5£  Ib. 

(27)  The  dimensions  of   a  return-tubular   boiler  are  as 
follows:  Diameter,  60  inches;  length  between  heads,  16  feet; 
outside  diameter  of  tubes,  3£  inches;  number  of  tubes,  64; 
distance  of  mean  water-line  from  top  of  boiler,  18  inches. 
(a)  Compute  the  steam   space  of  the  boiler  in  cubic  feet. 
(b}  Determine  the  number  of  gallons  of  water  that  will  be 
required  to  fill  the  boiler  up  to  the  mean  water  level. 

Ans    -f  ^     79"3  cu"  ft> 

I  (b)     1,246  gal.,  nearly. 

(28)  The  length  of  the  circumference  of  the  base  of  a 
cone    is   18.8496    inches  and  its  slant  height  is   10  inches. 
Find  the  area  of  the  entire  surface  of  the  cone. 

Ans.   122.5224  sq.  in. 

(29)  If  the  altitude  of  the  above  cone  were    9  inches, 
what  would  be  its  volume  ?  Ans.   84.8232  cu.  in. 

(30)  A  square  vat  is  11  feet  deep,  15  feet  square  at  the 
top,  and  12  feet  square  at  the  bottom.     How  many  gallons 
will  it  hold  ?  Ans.   15,058.29  gal. 

(31)  How  many  pails  of  water  would  be  required  to  fill 
the  vat,  the  pail  having  the  following  dimensions:  Depth, 


§  3  USE  OF  LETTERS  IN  FORMULAS.  5 

11  inches;  diameter  at  the  top,  12  inches;  diameter  at  the 
bottom,  9  inches  ?  Ans.   3,627.28. 

(32)  Find  (a)  the  area  of  the  surface,  and  (b)  the  cubical 
contents  of  a  ball  22£  inches  m  diameter. 

Ans    (  (a)     1,590.435  sq.  in. 
'  (  (b)     5,964.1313  cu.  in. 

(33)  (a)  What  is  the  volume   and  area   of  a  cylindrical 
ring  whose  outside  diameter  is  16  inches  and  inside  diameter 
13   inches  ?     (b}  If  made  of  cast  iron,   what  is  its  weight  ? 
Take  the  weight  of  1  cubic  inch  of  cast  iron  as  .261  pound. 

Ans.  Weight  =  21  Ib. 


PRINCIPLES  OF  MECHANICS. 


EXAMINATION  QUESTIONS. 


(1)  Define  mechanics. 

(2)  Define  (a)  matter;   (b)  molecule;  (c]  atom. 

(3)  In  what  three  states  does  matter  exist  ? 

(4)  Explain  clearly  the  distinction  between  general  prop- 
erties of  matter  and  special  properties  of  matter. 

(5)  Define  (a)  motion ;   (b)  velocity ;   (c)  uniform  velocity ; 
(d)  variable  velocity. 

(6)  Define  (a)  acceleration;  (b)  retardation;  (c)  average 
velocity. 

(7)  An  ocean  steamer  made  a  run  of  3,240  miles  in  6  days 
and   16  hours.     What  was  its  average  speed  in  miles  per 
hour  ?  Ans.  20£  mi. 

(8)  How  long  would  it  take  to  make  a  tour  around  the 
world  when  traveling  at  an  average  speed  of  3,000  feet  per 
minute,    assuming   the    distance    around,  the    world    to    be 
25,000  miles  ?  Ans.   30  da.  13  hr.  20  min. 

(9)  How  do  we  recognize  the  existence  of  a  force  ? 

(10)  What  conditions  are  necessary  to  'compare  the  rela- 
tive effects  of  different  forces  on  different  bodies  ? 

(11)  State  Newton's  Laws  of  Motion. 

(12)  What  do  you  understand  by  the  term  "inertia"? 

(13)  Define  (a)  dynamics;   (b)  statics. 

§4 


2  PRINCIPLES  OF  MECHANICS.  §  4 

(14)  Why  is  not  the  weight  of  a  given  body  the  same  at 
every  point  on  the  surface  of  the  earth  ? 

(15)  Determine    the    mass   of   a    body  that  weighs    346 
pounds  at  a  place  where  g  is  equal  to  32. 174.       Ans.   10. 75+. 

(16)  How  does  the  position  of  a  body  above  or  below 
the  surface  of  the  earth  affect  its  weight  ? 

(17)  A  locomotive  weighing  30  tons  has  to  overcome  a 
constant  force  of  15  pounds  per  ton  when  it  is  in  motion. 
What  total  force  must  the  locomotive  exert  so  that  its  speed 
may  increase  at  the  rate  of  3  feet  per  second  ? 

Ans.   6,047  Ib. 

(18)  What  will  be  the  momentum  of  the  locomotive  in  the 
preceding  example  when  it  has  attained  a  velocity  of  1  mile 
per  hour?  Ans.   2,736.3  Ib. 

(19)  Define  (a]  work;  (b)  power;  (c]  energy. 

(20)  What  horsepower  is  required  to  raise  a  .body  weigh- 
ing 66,000  pounds  through  a  distance  of  80  feet  in  \  hour  ? 

Ans.   5£  H.  P. 

(21)  A  body  weighing  6,432  pounds  is  moving  with  a 
constant  velocity  of  60  feet  per  second.     What  horsepower 
will  be  required  to  bring  the  body  to  rest  in  3  minutes  ? 

Ans.   3r7TH.  P. 

(22)  What  is  the  tangential  pressure  on  the  crank  of  an 
engine  when  the  crank  is  on  either  dead  center  ? 

(23)  What  is  friction  ? 

(24)  A  body  weighing  5,000  pounds  rests  on  a  horizontal 
surface.     In  order   to   slide  the  body  along  the  surface,  a 
horizontal  force  of  300  pounds  must  be  exerted.     What  is 
the  coefficient  of  friction  in  this  particular  case  ?     Ans.   .06. 

(25)  A   crosshead   weighing    1,000   pounds   and   having 
bronze  shoes  slides  on  a  slightly  greased,  horizontal  wrought- 
iron   guide.      If   the    contact   area   of  the   bronze   shoe   is 
100  square  inches,  what  will  be  (a)  the  total  friction  and  (b) 
the  friction  per  square  inch  of  contact  surface  ? 

(a)     160  Ib. 
1.6  Ib. 


§  4  PRINCIPLES  OF  MECHANICS.  3 

(20)     What  is  the  center  of  gravity  of  a  body  ? 

(27)  Explain  the  method  of  finding  the   common  center 
of   gravity  of    several    bodies  whose  weights    and    the  dis- 
tances between  whose  centers  of  gravity  are  known. 

(28)  Give  a  practical  method  of  determining  the  center 
of  gravity  of  a  solid  body. 

(29)  Define  (a)  centrifugal  force;  (b)  centripetal  force. 

(30)  What  is  the  relation  between  the  centrifugal  and 
centripetal  forces  of  a  revolving  body  ? 

(31)  A  body  weighing  10  pounds  revolves  at  a  speed  of 
60  revolutions  per  minute  about  a  point  6  feet  from  its  cen- 
ter of  gravity.     What   is  the   centrifugal   force   tending  to 
pull  the  body  from  the  point  about  which  it  revolves  ? 

Ans.   73.441b. 

(32)  Name    the    three    states    of    equilibrium    and   give 
examples  of  each. 

(33)  In  the  case  of  a  body  at  rest,  what  are  the  condi- 
tions of  the  forces  acting  on  that  body  ? 

(34)  How  is  the  condition   of  equilibrium   of  the  forces 
acting  on  a  body  affected  by   the   position  of  its  "line  of 
direction  "  with  respect  to  the  base  ? 


MACHINE  ELEMENTS. 


EXAMINATION  QUESTIONS. 

(1)  Define    (a)   lever;    (b)   weight   arm;    (c)   force    arm; 
(d)  fulcrum. 

(2)  What  is  the  condition  necessary  for  the  equilibrium 
of  the  lever  ? 

(3)  State   the   general  rule  that  expresses  the  relation 
existing  between  the  weight,  the  force,  and  the  distances 
through  which  they  move. 

(4)  What  must  be  the  length  of  the  weight  arm  in  order 
that  a  force  of  12  pounds  at  a  distance  of  20  inches  from  the 
fulcrum  will  raise  a  weight  of  100  pounds  at  the  end  of  the 
weight  arm  ?  Arts.   2.4  in. 

(5)  Into  what  two  classes  may  pulleys  be  divided  in  ref- 
erence to  their  construction  ? 

(6)  What  advantages  have  split  pulleys  over  solid  pul- 
leys ? 

(7)  Explain  how  a  crowned  pulley  tends  to  prevent  the 
slipping  off  of  the  belt. 

(8)  What  is  meant  by  "balancing  pulleys,"  and  how  is 
it  accomplished  ? 

(9)  Define  the  terms  "  driver  "  and  "  driven  "  as  applied 
to  pulleys. 

§5 

//.     8.     /.— 25 


2  MACHINE  ELEMENTS.  §  5 

(10)  Find  what  diameter  driver  will  be   required  which, 
when  running  at  600  revolutions  per  minute,  will  cause  the 
driven,  whose  diameter  is  6  inches,  to  run  at  1,800  revolu- 
tions per  minute.  Ans.   18  in. 

(11)  What  must  be  the  speed  of  a  driver  12  inches  in 
diameter  in  order  that  the  driven,  whose  diameter  is  5  inches, 
may  make  1,600  revolutions  per  minute  ? 

Ans.   666f  rev.  per  min. 

(12)  The  distance  between  the  centers  of  two  pulleys, 
whose  diameters  are  6  feet  and  2  feet,  respectively,  is  40  feet. 
What  is  the  required  length  of  an  open  belt  ?        Ans.   93  ft. 

(13)  A  single  belt  running  at   1,650  feet  per  minute- is 
used  to  transmit  40  horsepower.     If  the  arc  of  contact  on 
the  small  pulley  is  120°,  how  wide  should  the  belt  be  ? 

Ans.  28  in. 

(14)  What  horsepower  will  be  transmitted  by  the  belt 
in  question  13  if  the  velocity  is  reduced  to  1,200  feet  per 
minute?  Ans.   29.3  H.  P. 

(15)  If,  in  question  13,   a  double  belt  were  substituted 
for  the  single  belt,  how  wide  should  it  be  ?  Ans.   20  in. 

(16)  Which  side  of  a  leather  belt  should  be  in  contact 
with  the  pulley  face,  and  why  ? 

(17)  Should  rosin  be  used  on  a  belt  to  prevent  slipping  ? 

(18)  Give  the  causes  of  flapping  belts  and  their  rem- 
edies. 

(19)  State  the  different  methods  used  in  joining  belts. 
Which  of  these  makes  the  best  joint  ? 

(20)  What   precaution  should  be  observed  when  using 
rubber  belts  ? 

(21)  Define    the    terms     "driver"    and    "follower"    as 
applied  to  a  train  of  gear-wheels. 

(22)  In  Fig.  19,  if  the  diameter  of  A  is  90  inches,  that  of 
F  30  inches,  and  if  B  has  12  teeth,  C  30  teeth,  D  20  teeth, 
and  E  36  teeth,  find  the  weight  that  a  force  of  50  pounds  at 
/*can  raise.  Ans.   675  Ib. 


§  5  MACHINE  ELEMENTS.  3 

(23)  What  is  (a)  circular  pitch  ?  (b]  diametral  pitch  ? 

(24)  What  are  the  most  common  forms  of  teeth  used  in 
ordinary  practice  ? 

(25)  What  advantages  have  involute  teeth  over  epicy- 
cloidal  teeth  ? 

(26)  Find   the   pitch  diameter  of  a  gear-wheel  having 
60  teeth  and  a  circular  pitch  of  1.152  inches.         Ans.  22  in. 

(27)  What  is  the  circular  pitch  of  a  gear-wheel  30  inches 
in  diameter  having  60  teeth  ?  Ans.   1.57  in. 

(28)  What    is   the   over-all   diameter   of    a    gear-wheel 
having  80  teeth  with  a  diametral  pitch  of  8  ?       Ans.    10^-  in. 

(29)  Find  the  number  of  teeth  in  a   gear-wheel  whose 
outside  diameter  is  7£  inches  and  whose  diametral  pitch 
is  8.  Ans.   58  teeth. 

(30)  What  distinguishes  a  fixed  pulley  from  a  movable 
pulley  ? 

(31)  In  a  certain  combination  of  pulleys  there  are  seven 
movable  ones.     Neglecting  losses  due  to  friction,  how  heavy 
a  weight  can  a  force  of  150  pounds  raise  when  applied  to 
the  free  end  of  the  rope  ?  Ans.  2,100  Ib. 

(32)  How  does  friction  affect  the  force  required  to  raise 
a  given  weight  by  means  of  a  rope  or  chain  and  pulleys  ? 

(33)  In  an  ordinary  block  and  tackle   having  four  mov- 
able pulleys,  what  is  the  probable  actual  force  that  must  be 
applied  to  raise  a  weight  of  1,500  pounds  ?        Ans.  312.5  Ib. 

(34)  In  what  respect  is  the  Weston  differential  pulley 
block  better  than  the  ordinary  block  and  tackle  ? 

(35)  An  inclined  plane  is  70  feet  long  and  12  feet  high. 
What  force  acting  parallel  to  the  plane  will  be  required  to 
sustain  a  weight  of  600  pounds  on  the  plane  ?      Ans.   103  Ib. 

(36)  What  weight  will  a  force  of  36  pounds  acting  paral- 
lel to  the  base  of  the  plane  be  able  to  sustain  on  an  inclined 
plane  having  a  base  50  feet  long  and  a  height  of  15  feet  ? 

Ans.  120  Ib. 


1  MACHINE  ELEMENTS.  §  5 

(37)  What   is   the   probable  actual  weight  that  can  be 
raised  by  means  of  a  screw  jack  that  has  a  screw  2£  inches 
in  diameter  with  6  threads  to  the  inch  if  a  force  of  50  pounds 
is  applied  at  the  end  of  a  lever  20  inches  from  the  shaft  ? 

Ans.   2,356.2  Ib. 

(38)  Define  (a)  velocity  ratio;  (b}  efficiency. 

(39)  What   is  the  efficiency  of  the  screw  jack  of  ques- 
tion 37  ?  Ans.   6^  per  cent. 


MECHANICS  OF  FLUIDS. 


EXAMINATION  QUESTIONS. 

(1)  Define  hydrostatics. 

(2)  How  can  it  be  proved  that  liquids  transmit  pressure 
in  all  directions  and  with  the  same  intensity  ? 

(3)  State  Pascal's  law. 

(4)  In  what  direction  does  the  pressure  due  to  the  weight 
of  a  body  of  water  act  in  a  vessel  in  which  water  is  con- 
tained ? 

(5)  What  is  the  pressure  on  the  bottom  of  a  vessel  if  the 
base  is  a  circle  3  inches  in  diameter,  the  height  8  inches,  and 
the  vessel  is  completely  filled  with  water  ?         Ans.   2.045  Ib. 

(6)  State  the  law  for  the  upward  pressure  of  a  liquid  on 
a  horizontal  surface  submerged  in  the  liquid. 

(7)  Why  is  it  that  in  several  pipes  that  communicate  with 
one  another   and  that  differ  in  shape  and  size,  water  will 
stand  at  the  same  height  ? 

(8)  Why  is  it  that  water  issuing  from  a  hose  connected  to 
a  hydrant  cannot  be  made  to  spout  up  to  a  level  with  the 
surface   of    the   water   in   the   reservoir   that   supplies   the 
hydrant  ? 

(9)  Define  specific  gravity. 


2  MECHANICS  OF  FLUIDS.  §  6 

(10)  Calculate  the  weight  of  a  block  of  aluminum  whose 
volume  is  100  cubic  feet.  Ans.   15,605  Ib. 

(11)  State  the  principle  of  Archimedes. 

(12)  Explain  how,  by  the  application  of  the  principle  of 
Archimedes,  the  volume  of  an  irregularly  shaped  body  may 
be  accurately  determined. 

(13)  (a)  What    are    hydrometers  ?      (b)  Into   what    two 
classes  may  hydrometers  be  divided  ? 

(14)  A  single-cylinder  pump  feeds   a  boiler    through    a 
delivery  pipe  1  inch  in  actual  diameter.     The  piston  speed 
is  such  as  to  give  a  velocity  of  flow  of  400  feet  per  minute. 
How  many  gallons  of  water  can  be  pumped  into  the  boiler 
in  1  hour  ?  Ans.   979.2    gal. 

(15)  What  should  be  the  commercial  size  of  a  delivery 
pipe  from  a  duplex  pump  to  deliver  936  gallons  of  water  per 
hour  ?  Ans.   1  in. 

(16)  Why  is  it  important  to  have  as  few  bends  as  possible 
in  the  suction  pipe  leading  to  a  pump  ? 

(17)  What  simple  experiment  proves  that  gases  tend  to 
expand  and  increase  their  volume  ? 

(18)  How  high  a  column  of  mercury  will  the  pressure  of 
the  atmosphere  support  ? 

(19)  How  can  the  degree  of  the  vacuum  in  a  vessel  be 
determined  ? 

(20)  How  high  a  column  of  a  liquid  whose  specific  gravity 
is  2.5  will  the  atmospheric  pressure  support  ? 

Ans.   163.2  in. 

(21)  How  is  the  pressure  of  the  atmosphere  measured  ? 

(22)  Why  does  the  pressure  of  the  atmosphere  decrease 
as  the  elevation  above  sea  level  increases  ? 

(23)  In  what  respect  is  the  action  of  the  pressure  of  the 
atmosphere  similar  to  that  of  the  pressure  of  a  liquid  ? 


§  6  MECHANICS  OF  FLUIDS.  3 

(24)  How  does  the  tension  of  a   gas  change  with  the 
change  in  volume  under  constant  temperature  ? 

(25)  Define  (a)  gauge  pressure ;  (b)  absolute  pressure. 

(26)  A  metallic  tube  closed  at  one  end  is  fitted  with  an 
air-tight  movable  piston.     When  the  piston  is  at  the  open 
end  of  the  tube,  the  pressure  of  the  air  within  the  tube  is 
equal  to  14.7  pounds  per  square  inch.     What  will  be  the 
pressure  of  the  air  between  the  piston  and  the  closed  end  of 
the  tube  after  the  former  has  moved  toward  the  latter  a  dis- 
tance equal  to  ^  the  length  of  the  tube,  the  temperature 
remaining  constant  ?  Ans.  73.5  Ib.  per  sq.  in. 

(27)  Suppose  that  the  piston  in  the  tube  mentioned  in 
question  26  had  been  moved  toward  the  closed  end  until  the 
pressure  had  reached  147  pounds  per  square  inch.     What 
fraction  of  the  original  volume  would  the  compressed  air 
have  occupied  ?  Ans.  -fa. 

(28)  In  pneumatics,  what  is  an  air  pump  ? 

(29)  Can   a  perfect    vacuum   be  produced   with  the  air 
pump  ? 

(30)  Explain  the  operation  of  the  dashpot  of  a  Corliss 
engine. 

(31)  Explain  the  principle  of  operating  the  siphon. 

(32)  Suppose  that  two  vessels,  in  one  of  which  the  water 
stands  at  a  higher  level  than  in  the  other,  are  connected  by 
a  siphon  and  water  is  siphoned  from  the  vessel  in  which  it 
stands  at  the  higher  level  into  the  other  vessel.     What  will 
happen  when   the  water  in  each  vessel   reaches   the   same 
level  ? 

(33)  In  practice,  how  far  above  the  water  level  in  the 
vessel  from  which  water  is  siphoned  can  the  highest  point  of 
the  siphon  be  carried  ? 

(34)  (a)  What  is  a  pump  ?      (b)  Into   how  many  types 
may  pumps  be  divided  in  reference  to  their  mode  of  action  ? 
(c)  Name  these  types. 


4  MECHANICS  OF  FLUIDS.  §  6 

(35)  Explain  the  principle  on  which  the  suction  pump 
acts. 

(36)  What  advantage  has  a  lifting  pump  over  a  suction 
pump  ? 

(37)  In  what  respect  does  a  force  pump  differ  from  a 
lifting  pump  ? 

(38)  What  is  the  difference  between  a  single-acting  and 
a  double-acting  force  pump  ? 


STRENGTH  OF  MATERIALS. 


EXAMINATION  QUESTIONS. 

(1)  (a)   Define  stress,      (b)  Name    the    various   kinds   of 
stresses  to  which  a  body  can  be  subjected. 

(2)  A  weight  of  8,000  pounds  rests  on  the  top  of  a  cubi- 
cal block  of  wood  the  area  of  each  face  of  which  is  20  square 
inches.     What  is  the  unit  stress,  in  pounds  per  square  inch, 
to  which  the  block  is  subjected  ?          Ans.   400  Ib.  per  sq.  in. 

(3)  Define  (a)  strain;  (ft)  elasticity;  (c)  elastic  limit. 

(4)  Taking  the  ultimate  tensile  strength  of  wrought  iron 
as    55,000   pounds  per  square   inch,  what  tensile  force  will 
rupture   a   wrought-iron   bar   whose  cross-sectional  area   is 
4  square  inches  ?  Ans.   220,000  Ib. 

(5)  How  does  annealing  improve  old  chains  ? 

(6)  Give  a  practical  method  of  determining  the  true  con- 
dition  of   a  given    rope,   supposing   that   its   outer  surface 
appears  to  be  in  good  condition. 

(7)  (A)    On   what   does  the  safe    lifting   load  of   a  sling 
depend  ?     (V)  Explain  how  the  style  of  attachment  to  the 
hook  of  the  tackle  block  affects  the  amount  of  load  to  be 
raised. 

(8)  (a)  For  what  is  manila  rope  chiefly  used  ?     (b)  What 
is  the  object  of  lubricating  the  fibers  of  a  rope  ? 

§7 


2  STRENGTH  OF  MATERIALS.  §  7 

(9)  In  general,  how  should  the  size  of  the  pulley  compare 
with  the  size  of  manila  rope  used  for  transmitting  power  ? 

(10)  What  is  the  greatest  load  to  which  an  iron-wire  rope 
1£  inches  in  circumference  should  be  subjected  ? 

Ans.   1,350  Ib. 

(11)  What  should  be  the  circumference  of  a  steel-wire  rope 
under  a  maximum  working  load  of  16,000  pounds  ? 

Ans.   4  in.,  nearly. 

(12)  How  does  the  strength  of  a  column  having  both  ends 
flat  compare  with  that  of  columns  both  of  whose  ends  are  not 
flat  but  which  in  every  other  respect  are  similar  to  the  first 
column  ? 

(13)  Give  practical  examples  of  the  three  different  classes 
into  which  columns  are  divided  with   respect  to  the  condi- 
tion of  their  ends. 

(14)  What  is  the  safe  steady  working  load  that  a  cast- 
iron  column,  having  fixed  ends,  14  inches  in  diameter  and 
16  feet  high  can  sustain  ?  Ans.   1,291,480  Ib. 

(15)  Would   you  consider  a  steel  piston  rod  6  inches  in 
diameter  of  sufficient  size  for  a  40-inch  cylinder  using  steam 
at  110  pounds  pressure  ?     If  so,  why  ? 

(16)  A  solid  yellow-pine  beam  14  inches  square  rests  on 
two  supports  12  feet  apart.     What  steady  safe  load  will  the 
beam  support  at  its  middle  point  ?  Ans.   18,547  Ib. 

(17)  An  oak  beam   4  inches  wide  and  6  inches  deep  is 
subjected  to  a  sudden  shearing  stress  of  10,000  pounds  across 
the  grain.     Is  this  a  safe  load  for  the  given  conditions  ? 

(18)  What  should  be  the  area  of  a  wrougJit-iron  beam  to 
safely  support  a  sudden  shearing  stress  of  400,000  pounds  ? 

Ans.   91  sq.  in.,  nearly. 

(19)  What  do  you  understand  by  double  shear  ? 

(20)  What  is  the  use  of  countershafts  ? 

(21)  What  is  the  distinction  between  cold-rolled  shafting 
and  bright  shafting  ? 


§  7  STRENGTH  OF  MATERIALS.  3 

(22)  How   is   bright    shafting   designated    commercially 
with  respect  to  size  ? 

(23)  Why  is  it  good  practice  to  place  pulleys  for  trans- 
mitting or  receiving  power  as  near  the  bearings  of  the  shaft 
as  possible  ? 

(24)  What  horsepower  can  be  transmitted  from  a  steel 
shaft  8  inches  in  diameter  by  means  of  pulleys  between  the 
bearings  when    running  at   a   speed  of   80   revolutions    per 
minute  ?  Ans.  482  H.  P. 

(25)  What  must  be  the  speed  of  a  4-inch  shaft  of  cold- 
rolled  iron,  having  no  pulleys  between  bearings,  to  transmit 
75  horsepower  ?  Ans.   76  rev. 

(26)  Find  the  diameter  of  a  wrought-iron  shaft  running 
at  300  revolutions  per  minute  to  transmit  84  horsepower  by 
means  of  pulleys  between  its  bearings.  Ans.   3  in. 

(27)  How  does  a  change  in  the  speed  of  a  shaft  affect  the 
amount  of  power  transmitted  ? 

(28)  Does  a  high  tensile  strength  in  metals  necessarily 
imply  an  ability  on  the  part  of   the  metal  to  safely  resist 
repeated  applications  of  sudden  stresses  ? 

(29)  What  is  the  chief  advantage  of  a  steel  rope  over  an 
iron  rope  ? 


A  KEY 

TO    ALL    THE 

QUESTIONS    AND    EXAMPLES 

CONTAINED    IN    THE 

EXAMINATION    QUESTIONS 

INCLUDED  IN  THIS  VOLUME. 


The  Keys  that  follow  have  been  divided  into  sections  cor- 
responding to  the  Examination  Questions  to  which  they 
refer,  and  have  been  given  corresponding  section  numbers. 
The  answers  and  solutions  have  been  numbered  to  corre- 
spond with  the  questions.  When  the  answer  to  a  question 
involves  a  repetition  of  statements  given  in  the  Instruction 
Paper,  the  reader  has  been  referred  to  a  numbered  article, 
the  reading  of  which  will  enable  him  to  answer  the  question 
himself. 

To  be  of  the  greatest  benefit,  the  Keys  should  be  used 
sparingly.  They  should  be  used  much  in  the  same  manner 
as  a  pupil  would  go  to  a  teacher  for  instruction  with  regard 
to  answering  some  example  he  was  unable  to  solve.  If  used 
in  this  manner,  the  Keys  will  be  of  great  help  and  assist- 
ance to  the  student,  and  will  be  a  source  of  encouragement 
to  him  in  studying  the  various  papers  composing  the  Course. 


ARITHMETIC. 

(PART  1.) 


(1)  See  Art.  1. 

(2)  See  Art.  3. 

(3)  See  Arts.  5  and  6. 

(4)  See  Arts.  1O  and  11. 

(5)  980  =  Nine  hundred  eighty. 
605  =  Six  hundred  five. 

28,284  —  Twenty-eight  thousand  two  hundred  eighty-four. 
9,006,042  =  Nine  million  six  thousand  forty-two. 
850,317,002  =  Eight  hundred  fifty  million  three  hundred 
seventeen  thousand  two. 

700, 004  —  Seven  hundred  thousand  four. 

(6)  Seven  thousand  six  hundred  =  7,600. 
Eighty-one  thousand  four  hundred  two  =  81,402. 
Five  million  four  thousand  seven  =  5,004,007. 

One  hundred  and  eight  million  ten  thousand  one  =  108,- 
010,001. 

Eighteen  million  six  =  18,000,006. 
Thirty  thousand  ten  =  30,010. 
§1 

For  notice  of  copyright,  see  page  immediately  following  the  title  page. 


2  ARITHMETIC.  §  1 

(7)  In  adding  whole  numbers,    place  the  numbers  to  be 
added  directly  under  each  other,  so  that  the  extreme  right- 
hand  figures  will  stand   in  the  same 

3290  column,  regardless  of  the  position  of 

504  those    at    the    left.     Add    the   first 

865403  column  of   figures    at    the    extreme 

2074  right,    which    equals   19   units,  or  1 

ten  and   9  units.     We   place  9  units 

7  under  the  units  column  and  reserve 

871359     Ans.     1  ten  for  the  column  of  tens,  8  +  7 

+  9  +  1  =  25  tens,  or  2  hundreds  and 

5  tens.  Place  5  tens  under  the  tens  column  and  reserve 
2  hundreds  for  the  hundreds  column.  4  +  5+2  +  2=:  13 
hundreds,  or  1  thousand  and  3  hundreds.  Place  3  hundreds 
under  the  hundreds  column  and  reserve  the  1  thousand  for 
the  thousands  column.  2  +  5+3  +  1  =  11  thousands,  or 
1  ten-thousand  and  1  thousand.  Place  the  1  thousand  in 
the  column  of  thousands  and  reserve  the  1  ten-thousand 
for  the  column  of  ten-thousands.  6  +  1  =  7  ten-thousands. 
Place  this  7  ten-thousands  in  the  ten-thousands  column. 
There  is  but  one  figure,  8,  in  the  hundreds  of  thousands 
place  in  the  numbers  to  be  added,  so  it  is  placed  in  the 
hundreds  of  thousands  column  of  the  sum. 

A  simpler  (though  less  scientific)  explanation  of  the  same 
problem  is  the  following :  7  +  1  +  4+3  +  4  +  0  =  19;  Avrite 
the  9  and  reserve  the  1.  8  +  7  +  0  +  0+9  +  1  reserved 
=  25 ;  write  the  5  and  reserve  the  2.  0  +  4  +  5  +  2+2 
reserved  =  13 ;  write  the  3  and  reserve  the  1.  2  +  5  +  3+1 
reserved  =  11 ;  write  the  1  and  reserve  1.  6  +  1  reserved  =  7 ; 
write  the  7.  Bring  down  the  8  to  its  place  in  the  sum. 

(8)  709 
8304725- 

391 

100302 
300 
909 


8407336  Ans. 


§  1  ARITHMETIC.  3 

(9)  The  steam  engine,  during  the  12-hour  test,  showed 
that  the  number  of  revolutions  made  were  150,508,  since 
12,600  +  12,444+  12,467  +  12,528  +  12,468  +  12,590  +  12,610 
+  12,589  +  12,576  +  12,558  +  12,546  +  12,532  =  150,508  rev. 

12600  revolutions. 
12444  revolutions. 

12467  revolutions. 
12528  revolutions. 

12468  revolutions. 
12590  revolutions. 
12610  revolutions. 
12589  revolutions. 
12576  revolutions. 
12558  revolutions. 

12546  revolutions. 

/ 

12532  revolutions. 


150508  revolutions.     Ans. 

(10)     In  subtracting  whole  numbers,  place  the  subtrahend, 

or  smaller  number,  under  the  minuend,  or  larger  number, 

(a]     5  0  9  6  2  so  that  t^ie  right-hand  figures  stand 

3  3  3  g  directly  under  each  other.     Begin  at 

the   right   to    subtract.     We  cannot 

47624     Ans.   subtract  g  units  from  2  units,  so  we 

take  1  ten  from  the  6  tens  and  add  it  to  the  2  units.     1  ten 

=  10  units,  so  we  have  10  units  +  2  units  —  12  units.     Then, 

8  units  from  12  units  leaves  4  units.     We  took  1  ten  from 
6  tens,  so  only  5  tens  remain.     3   tens  from   5  tens  leaves 
2  tens.      In  the  hundreds  column  we  have  3 -hundreds   from 

9  hundreds  leaves  6  hundreds.     We  cannot  subtract  3  thou- 
sands from  0   thousands,    so  we  take  1   ten-thousand  from 
5  ten-thousands  and  add  it  to  the  0  thousands.     1  ten-thousand 
=  10  thousands,  and  10  thousands  +  0  thousands  =  10  thou- 
sands.     Subtracting,   we  have   3  thousands   from   10  thou- 
sands leaves   7  thousands.     We  took  1   ten-thousand  from 
5  ten-thousands  and  have  4  ten-thousands  remaining.      Since 
there  are  no  ten-thousands  in  the  subtrahend,  the  4  in  the 

H.    s.    I.— 26 


4  ARITHMETIC.  §  1 

ten-thousands  column  in  the  minuend  is  brought  down  into 
the  same  column  in  the  remainder,  because  0  from  4  leaves  4. 

(b)  15339 
10001 


5338  Ans. 

(11)  (a)  7  0  9  6  8        (b)  I  0  0  0  0  0 
32975  98735 


37993     Ans.  1265     Ans. 

(12) 
3040  =  No.  of  gallons  in  the  tank  at  the  beginning  of  the 

day. 

4780=  No.  of  gallons  pumped  in  during  the  morning. 
7820  =  No.  of  gallons  in  the  tank  after  4,780  gallons  were 

added. 

7240  =  No.  of  gallons  drawn  out  during  the  morning. 
580  =  No.  of  gallons  in  the  tank  at  the  beginning  of  the 

afternoon. 

8675=  No.  of  gallons  pumped  in  during  the  afternoon. 
9255  =  No.  of  gallons  in  the  tank  after  8,675  gallons  were 

added. 

7633  =  No.  of  gallons  drawn  out  during  the  afternoon. 
1622=  No.  of  gallons  remaining  in  the  tank  at  night.  Ans. 

(13)  In  the  multiplication  of  whole  numbers,  place  the 
multiplier  under  the  multiplicand  and  multiply  each  term 
of  the  multiplicand  by  each  term  of  the  multiplier,  writing 
the  right-hand  figure  of  each  product  obtained  under  the 
term  of  the  multiplier  which  produces  it. 

(a)        526387  7  times  7  units  =  49  units,  or 

7  4  tens  and  9  units.     We  write 

3684709     Ans       the    9    units    and    reserve    the 

4  tens.   7  times 8  tens  =  56  tens; 

56  +  4  tens  reserved  =  60  tens,  or  6  hundreds  and  0  tens. 
Write  the  0  tens  and  reserve  the  6  hundreds.  7x3  hun- 
dreds =  21  hundreds;  21  +  6  hundreds  reserved  =  27  hun- 
dreds, or  2  thousands  and  7  hundreds.  Write  the  7  hun- 
dreds and  reserve  the  2  thousands.  7x6  thousands  = 


§  1  ARITHMETIC.  5 

42  thousands;  42  +  2  thousands  reserved  =  44  thousands, 
or  4  ten-thousands  and  4  thousands.  Write  the  4  thousands 
and  reserve  the  4  ten-thousands.  7x2  ten-thousands 
—  14 ten-thousands;  14  +  4  ten-thousands  reserved  =  18  ten- 
thousands,  or  1  hundred-thousand  and  8  ten-thousands. 
Write  the  8  ten-thousands  and  reserve  the  1  hundred- 
thousand.  7  X  5  hundred-thousands  =35  hundred-thousands; 
35  +  1  hundred-thousand  reserved  =  36  hundred-thousands. 
Since  there  are  no  more  figures  in  the  multiplicand  to  be 
multiplied,  we  write  the  36  hundred-thousands  in  the  prod- 
uct. This  completes  the  multiplication. 

A  simpler  (though  less  scientific)  explanation  of  the  same 
problem  is  the  following: 

7   times  7  =  49 ;  write  the  9  and  reserve  the  4.     7  times 

8  =  56;  56  +  4  reserved  =  60;  write  the  0  and  reserve  the  6. 
7    times    3  =  21;    21  +  6    reserved  =  27 ;  '  write   the    7    and 
reserve  the  2.     7  X  6  =  42;  42  +  2  reserved  =  44;  write  the 
4  and  reserve  4.     7  X  2  =  14;  14  +  4  reserved  =  18;  write 
the   8   and   reserve   1.       7  X  5  =  35;    35  +  1  reserved  =  36; 
write  the  36. 

In   this   case  the  multiplier  is 

(b)      7  0  0  2  9  8  17  units,  or  1  ten  and  7  units,  so 

17 

that  the  product  is  obtained  by 

4902086  adding     two     partial      products, 

700298  namely,     7  X  700,298    and    10  X 

11905066    Ans.     700^98      The  actual  operation  is 

performed  as  follows: 
7  times  8  =  56;  write  the  6   and  reserve  the  5.     7  times 

9  =  63;  63  +  5  reserved  =  68;  write  the  8  and  reserve  the  6. 
7   times  2  =  14;    14+6   reserved  =  20;    write    the    0    and 
reserve    the    2.     7    times  0  =  0;    0  +  2   reserved  =  2 ;  write 
the    2.      7   times  0  =  0;    0+0   reserved  =0;    write  the    0. 
7  times  7  =  49 ;  49  +  0  reserved  =  49 ;  write  the  49. 

To  multiply  by  the  1  ten  we  have  1  ten  times  8  units 
=  8  tens  and  0  units.  We  do  not  write  the  0  units,  but  write 
the  8  tens  under  the  8  tens  in  the  first  partial  product  above. 
We  next  multiply  1  ten  and  9  tens  =  90  tens  =  9  hun- 
dreds +  0  tens;  write  the  9  hundreds  under  the  0  hundreds 


6  ARITHMETIC.  §  1 

of  the  first  partial  product  above.  Again,  1  ten  times 
2  hundreds  —  2  thousands;  write  the  2  thousands  under  the 
2  thousands  above.  1  ten  times  0  thousands  =  0  ten- 
thousands;  write  the  0  in  the  ten-thousands  place.  1  ten 
times  0  ten-thousands  =  0  hundred-thousands;  write  the  0 
in  the  hundred-thousands  place.  1  ten  times  7  hundred- 
thousands  =  7  millions;  write  the  7  in  the  millions  place. 
This  completes  the  second  partial  product.  Add  the  two 
partial  products;  their  sum  equals  the  entire  product. 

(c)  217  Multiply    any   two  of    the    numbers 

103         together,   and   multiply  their  product 
jj~jj~J        by  the  third  number. 

000 

217 


22351 

6  7 

156457 
134106 

1497517  Ans. 

(14)  If  your  watch  ticks  every  second,  to  find  how  many 
times  it  ticks  in  1  week,  it  is  necessary  to  find  the  number  of 
seconds  in  1  week. 

6  0  seconds  =  1  minute. 

6  0  minutes  =  1  hour. 


0  0  seconds  =  1  hour. 
2  4  hours  =  1  day. 


14400 

7200 


86400  seconds  =  1  day. 
7  days  —  1  week. 

604800  seconds  in  1  week,  or  the  number  of  times  that 
your  watch  ticks  in  a  week.     Ans. 

(15)  (a)  If  an  engine  and  boiler  are  worth  $3,246,  and 
the  building  is  worth  three  times  as  much,  plus  $1,200,  then 
the  building  is  worth 


§  1  ARITHMETIC. 


$10938  =  value  of  building. 

If  the  tools  are  worth  twice  as  much  as  the  building,  plus 
$1,875,  then  the  tools  are  worth 

10938 


21876 
plus        1875 

$23751  =  value  of  tools. 

Value  of  building  =      10938 
Value  of  tools         =      23751 

$34689  =  value  of   the  building 

...      T,  t        r        .  and  tools.     Ans. 

(b)     Value  of  engine 

and  boiler  =          3246 
Value  of  build- 
ing and  tools  =      34689 

$37935  =  value     of     the    whole 
plant.     Ans. 

(16)  (a)  84)96284  2.0  000(1146  2.4  047+ 

4  Ans. 


400 


640 

588 

5  2 


8  ARITHMETIC.  §  1 

84  is  contained  once  in  96.  Place  1  as  the  first  figure  in 
the  quotient  and  multiply  the  divisor  84  by  it.  Subtract 
the  product,  which  is  84,  from  96,  leaving  a  remainder  of 
12.  Bring  down  the  next  figure  in  the  dividend,  which  is  2, 
and  annex  it  to  12,  making  a  new  dividend  of  122. 

84  is  contained  in  122  once.  Place  1  as  the  second  figure 
in  the  quotient  and  multiply  the  divisor  84  by  it.  Subtract 
the  product  (84)  from  122,  leaving  a  remainder  of  38.  Bring 
down  the  next  figure  in  the  dividend,  which  is  8,  and  annex 
it  to  38,  making  a  new  dividend  of  388. 

84  is  contained  in  388  4  times.  Place  4  as  the  third  figure 
in  the  quotient  and  multiply  the  divisor  84  by  it.  The  prod- 
uct is  336.  Subtract  the  product  from  388,  leaving  a 
remainder  of  52.  Bring  down  the  next  figure,  which  is  4, 
and  annex  it  to  52,  making  a  new  dividend  of  524. 

84  is  contained  in  524  6  times.  Place  6  as  the  fourth  figure 
in  the  quotient.  Multiply  the  divisor  84  by  it  and  subtract 
the  product  (504)  from  524,  leaving  a  remainder  of  20.  Bring 
down  the  next  figure,  which  is  2,  and  annex  it  to  20,  making 
a  new  dividend  of  202. 

84  is  contained  in  202  2  times.  Place  2  as  the  fifth  figure 
in  the  quotient.  Multiply  the  divisor  84  by  it  and  subtract 
the  product  (168)  from  202,  leaving  a  remainder  of  34.  :  If 
it  is  desired  to  carry  the  quotient  to  4  decimal  places,  annex 
4  ciphers  to  the  dividend  and  continue  in  the  same  way.  In 
the  quotient  point  off  as  many  decimal  places  as  there  are 
ciphers  annexed,  or,  in  this  case,  4  decimal  places. 

(b)          63)39728.000(630.603+     Ans. 
3  7 


§  1  ARITHMETIC.  9 

*63  is  not  contained  in  38,  so  we  place  a  cipher  in  the 
quotient  and  bring  down  the  next  figure  in  the  dividend, 
which  is  a  cipher  that  has  been  annexed,  making  a  new  divi- 
dend of  380.  63  is  contained  in  380  6  times.  6  X  63  =  378. 
Subtracting  378  from  380  leaves  2.  Bringing  down  the 
next  figure  in  the  dividend,  we  have  20  for  a  new  dividend. 
63  is  not  contained  in  20,  so  we  place  a  cipher  in  the  quo- 
tient and  bring  down  the  next  cipher  in  the  dividend,  making 
a  new  dividend  of  200.  63  is  contained  3  times  in  200. 

(c)  108)29714.0000(275.1296  Ans. 
216 


811   (d)  135)406089.0000(3008.0666 
756          405  Ans. 


554  1089* 

540  1080 


140  900 

108  810 


320  900 

216  810 


1040  900 

972  810 


680  90 

648 

~32 

*  135  is  not  contained  in  10,  so  we  place  a  cipher  as  the 
second  figure  in  the  quotient.  Bringing  down  the  next 
figure,  8,  and  annexing  it  to  10,  we  have  a  new  dividend  of 
108.  135  is  not  contained  in  108,  so  we  place  a  cipher  as 
the  third  figure  in  the  quotient  and  bring  down  the  next 
figure  in  the  dividend,  or  9,  and  annex  it  to  108,  making  a 
new  dividend  of  1,089.  135  is  contained  in  1,089  8  times. 
Write  8  as  the  fourth  figure  in  the  quotient.  Multiply  135 
by  8  and  subtract  the  product  (1,080)  from  1,089,  which 
leaves  a  remainder  of  9.  Bring  down  the  next  figure  in  the 
dividend,  which  is  a  cipher  that  has  been  annexed,  and  annex 


10  ARITHMETIC.  §  1 

it  to  the  remainder  9,  making  a  new  dividend  of  90.  As  135 
is  not  contained  in  90,  we  place  a  0  in  the  quotient  and 
bring  down  another  cipher  from  the  dividend,  making  a  new 
dividend  of  900.  135  is  contained  in  900  6  times.  Write  6 
as  the  next  figure  in  the  quotient,  multiply  135  by  6,  and 
subtract  the  product  (810)  from  900,  which  leaves  a 
remainder  of  90.  Bring  down  the  next  figure  (0)  in  the 
dividend  and  annex  it  to  the  remainder  90,  making  a  new 
dividend  of  900.  135  is  contained  in  900  6  times.  Place  6 
as  the  next  figure  in  the  quotient  and  multiply  the  divisor 
by  it.  It  is  plain  that  each  succeeding  figure  of  the  quotient 
will  be  6.  Point  off  four  decimal  places  in  the  quotient, 
since  four  ciphers  were  annexed. 

(17)  If  in  1  hour  10  pounds  of  coal  are  burned  for  every 
square  foot  of  grate  area  and  9  pounds  of  water  are  evapo- 
rated for  every  pound  of  coal  burned,  then  in  1  hour  there 
would  be  9  X  10  or  90  pounds  of  water  evaporated  for  1  sq.  ft. 
of  grate  area;    and  since  the  grate  area  is  30   sq.  ft.,  the 
amount  of  water  evaporated  would  be  30  X  90  =  2,700  Ib. 
Since  2,700  Ib.  of  water  are  evaporated  in  1   hour,  in  a  day 
of  10  hours  there  would  be  10  X  2,700  Ib.,  or  27,000  Ib.  of 
water  evaporated. 

(18)  If  a  mechanic  earns  $1,500  a  year  and  his  expenses 
are  $968   per   year,    then  he  would  save  $1500  —  $968,  or 
$532  per  year. 

1500 
968 


$532 

If  he  saves  $532  in  1  year,  to  save  $3,724  it  would  take  as 
many  years  as  $532  is  contained  times  in  $3,724,  or  7  years. 

532)3724(7  years.     Ans. 
3724 


,    (19)     (a)  (72  X  48  X  28  X  5)  -f-  (84  X  15  X  7  X  6). 

Placing  the  numerator  over  the  denominator,  the  problem 
becomes 


§  1  ARITHMETIC.  11 

72  X  48  X  28  X  5  _ 
84  X  15  X  7  X  6  ~ 

The  5  in  the  numerator  and  15  in  the  denominator  are 
both  divisible  by  5,  since  5  divided  by  5  equals  1  and  15 
divided  by  5  equals  3.  Cross  off  the  5  ;  also  cross  off  the 
15  and  write  the  3  under  it.  Thus, 

72  x  48  X  28  X  ft  _ 
84  X  #  X  7  X  6  ~ 
3 

72  in  the  numerator  and  84  in  the  denominator  are  divisible 
by  12,  since  72  divided  by  12  equals  6  and  84  divided  by  12 
equals  7.  .  Cross  off  the  72  and  write  the  6  over  it  ;  also, 
cross  off  'the  84  and  write  the  7  under  it.  Thus, 

6 

JT?  X  48  X  28  x  ft  _ 
ftl  X  tf  X  7  X  6  ~ 
7  3 

Again,  28  in  the  numerator  and  7  in  the  denominator  are 
divisible  by  7,  since  28  divided  by  7  equals  4  and  7  divided 
by  7  equals  1.  Cross  off\h&  28  and  write  the  4  0wr  it;  also, 
cross  off  titol.  Thus, 

6  4 

X  48  x  ffi  X  ft  _ 


Again,  48  in  the  numerator  and  6  in  the  denominator  are 
divisible  by  6,  since  48  divided  by  6  equals  8  and  6  divided  by 
6  equals  1.  Cross  off  the  48  and  write  the  8  over  it  ;  also, 
cross  off  \b&§.  Thus, 

684 


p£  X  lp  X  J  X  p 

Again,  6  in  the  numerator  and  3  in  the  denominator  are 
divisible  by  3,  since  6  divided  by  3  equals  2  and  3  divided  by 
3  equals  1.  Cross  offihz  6  and  write  the  2  over  it ;  also,  cross 
3.  Thus, 


12  ARITHMETIC. 

2 

8        4 
x  48  x  28  x 


jfy*  x  J0  x  /T  x 

7       0 

Since  there  are  «0  /w#  remaining  numbers  (one  in  the 
numerator  and  one  in  the  denominator)  divisible  by  any 
number  except  1,  without  a  remainder,  it  is  impossible  to 
cancel  further. 

Multiply  all  the  tincanceled  numbers  in  the  numerator 
together  and  divide  their  product  by  the  product  of  all 
the  uncanceled  numbers  in  the  denominator.  The  result 
will  be  the  quotient.  The  product  of  all  the  uncanceled 
numbers  in  the  numerator  equals  2  X  8  X  4  =  64;  the  product 
of  all  the  uncanceled  numbers  in  the  denominator  equals  7. 

2 


__ 

LC6'     #  x  tf  X  7  X  0  ~     ~T~         7  ~ 
7       ? 

(0)     (80  X  60  X  50  X  16  X  14)  •*•  (70  X  50  X  24  X  20). 
Placing  the  numerator  over  the  denominator,  the  problem 


becomes 


80  X  60  X  50  X  16  X  14       ? 


70  X  50  X  24  X  20 

The  50  in  the  numerator  and  70  in  the  denominator  are 
both  divisible  by  10,  since  50  divided  by  10  equals  5  and  70 
divided  by  10  equals  7.  Cross  off  the  50  and  write  the  5 
over  it  ;  also,  cross  off  the  70  and  write  the  7  under  it.  Thus, 

5 
80  X  60  X  00  X  16  X  14 

70  X  50  x  24  X  20 

7 

Also,  80  in  the  numerator  and  20  in  the  denominator  are 
divisible  by  20,  since  80  divided  by  20  equals  4  and  20 
divided  by  20  equals  1.  Cross  off  the  80  and  write  the  4  over 
it;  also,  cross  off  the  20.  Thus, 


§  1  ARITHMETIC.  13 

4  5 

0  X  60  x  W  x  16  x  14 


W 


X  50  x  24  x 


Again,  16  in  the  numerator  and  24  in  the  denominator  are 
divisible  by  8,  since  16  divided  by  8  equals  2  and  24  divided 
by  8  equals  3.  Cross  off  the  16  and  write  the  2  over  it  ; 
also,  cross  off  the  24  and  write  the  3  under  it.  Thus, 

4  52 

$0  X  60  X  ?0  X  ;0  X  14  _ 

W  X  50  X  ?4  X  20 
7  3 

Again,  60  in  the  numerator  and  50  in  the  denominator 
are    divisible  by  10,   since    60   divided   by  10   equals   6  and 
50  divided  by  10  equals  5.      Cross  off  the  60  and  write  the  6 
over  it  ;  also,  cross  off  the  50  and  write  the  5  under  it.    Thus, 
4652 

$0  x  00  x  w  x  ;0  x  14 


;0 
7 


x  0p  x  ?4  x  2 
53 


The  14  in  the  numerator  and  7  in  the  denominator  are 
divisible  by  7,  since  14  divided  by  7  equals  2  and  7  divided  by 
7  equals  1.  Cross  off  the  14  and  write  the  2  over  it  ;  also, 
cross  off  the  7.  Thus, 

46522 


;0 


x  £0  x  w  x  20 


The  5  in  the  numerator  and  5  in  the  denominator  are 
divisible  by  5,  since  5  divided  by  5  equals  1.      Cross  off  the 
5  of  the  dividend;  also,  cross  off  the  5  of  the  divisor.     Thus, 
46022 


The  6  in  the  numerator  and  3  in  the  denominator  are 
divisible  by  3,  since  6  divided  by  3  equals  2,  and  3  divided 
by  3  equals  1.  Cross  off  the  6  and  place  2  over  it;  also, 
cross  off  the  3.  Thus, 


14  ARITHMETIC.  §  1 


2        2 

;$  x  #  _ 


x?0 


4        0        £        2        2 
Hence,  ^  *  ^  X  ^  *  ?  *fl  ^  =  4  X  2  X  2  X  2  =  32. 


jf        ^        £  Ans. 

(20)     If  the  freight  train  ran  365  miles  in  one  week,  and 

3  times  as  far,  lacking  246  miles  the  next  week,  then  it  ran 

(3  X  365  miles)  —  246  miles,  or  849  miles  the  second  week. 

Thus, 

365 
3 


1095 
-    246 


849  miles.     Ans. 

(21)  The  distance  from  Philadelphia  to  Pittsburg  is 
354  miles.  Since  there  are  5,280  feet  in  1  mile,  in  354  miles 
there  are  354  X  5,280  feet,  or  1,869,120  feet.  If  the  driving 
wheel  of  the  locomotive  is  16  feet  in  circumference,  in  going 
from  Philadelphia  to  Pittsburg,  a  distance  of  1,869,120  feet, 
it  will  make  1,869,120  -=-  16,  or  116,820  revolutions. 

16)1869120(116820  rev.     Ans. 
1  6 


ARITHMETIC. 

(PART  2.) 


(1)  See  Art.  1. 

(2)  See  Art.  6. 

(3)  See  Art.  3. 

(4)  See  Art.  3. 

(5)  y  is  an  improper  fraction,  since  its  numerator  13  is 
greater  than  its  denominator  8. 

(6)  4£;  14T37;  85TV 

(7)  To  reduce  a  fraction  to  its  lowest  terms  means  to 
change  its  form  without  changing  its  value.     In  order  to 
do  this,  we  must  divide  both  numerator  and  denominator  by 
the  same  number  until  we  can  no  longer  find  any  number 
(except  1)  which  will  divide  both  of  these  terms  without  a 
remainder. 

To  reduce  the  fraction  f-  to  its  lowest  terms,  we  divide 
both    numerator  and   denominator    by    4  and  obtain   as   a 

4  -1-  4  4-^4 

result    the    fraction    £.     Thus,  -          =  \\    similarly,  — 

o  ~T~  4  J.O~r~  4 

j3_-^4_2-^-2  ^-^8_4H-4_1 

*'  33-=-4~~8-=-2~    ''   64-f-8~8-h4~~ 

(8)  When  the  denominator  of  any  number  is  not  expressed, 
it  is  understood  to  be  1,  so  that  f  is  the  same  as  6  -4-  1,  or6. 

§1 

For  notice  of  copyright,  see  page  immediately  following  the  title  page. 


16  ARITHMETIC.  §  1 

To  reduce  -f-  to  an  improper  fraction  whose  denominator  is 
4,  we  must  multiply  both  numerator  and  denominator  by 
some  number  which  will  make  the  denominator  of  6  equal 
to  4.  Since  the  denominator  is  1,  by  multiplying  both 

6x4 

terms  of  -f-  by  4,  we  will  have  -          =  -2¥4-,  which  has  the  same 

1X4 

value  as  6,  but  has  a  different  form. 

(9)  In  order  to  reduce  a  mixed  number  to  an  improper 
fraction,  we  must  multiply  the  whole  number  by  the  denomi- 
nator of  the  fraction  and  add  the  numerator  of  the  fraction 
to  the  product.  This  result  is  the  numerator  of  the  improper 
fraction,  of  which  the  denominator  is  the  denominator  of  the 
fractional  part  of  the  mixed  number. 

7£  means  the  same  as  7  -f-  £.  In  1  there  are  -f;  hence,  in 
7  there  are  7  X  f  =  -*-/-.  Add  the  |  of  the  mixed  number 
and  we  obtain  -%6-  +  1  —  T3-?  which  is  the  required  improper 
fraction. 

OiXM+«  .  (10X+8      . 


.. 

(10)     Each  stroke  of  the  engine  is  18  inches  in  length. 
Since  the    piston  makes  2  strokes  for  each  revolution,    it 
would  pass  over  a  distance  of   2  X  18  inches  =  36  inches, 
or  3  feet,    in    1    minute,    and    in   making 
480      ft.       leo  revolutions  it  would  pass  over  160  X  3, 
X        60  or  480  ft.     Since  480  ft.  are  passed  over 

2  8  8  0  0  ft.  m  1  minute,  in  1  hour,  or  60  minutes, 
the  distance  passed  over  equals  60  X  480 
=  28,800  ft.  Since  the  steam  engine  runs 
"  days  a  week  and  8£  hours  per  day,  the 


28800  total  number  of  hours  it   runs   per  week 

144000  =  6  X  8£,    or    51    hours.       If   the    piston 

1468800  passes    over  a    distance    of   28,800  ft.    in 

1   hour,  in   51   hours  it  would  pass  over 

51  X  28,800  ft.,  or  1,468,800  ft.     Ans. 


(11)     i+|  +  |  =  =  |=l.     Ans. 


§  1  ARITHMETIC.  17 

When  the  denominators  of  the  fractions  to  be  added  are 
alike,  we  know  that  the  units  are  divided  into  the  same 
number  of  parts  (in  this  case  eighths)  ;  we  therefore  add  the 
numerators  of  the  fractions  to  find  the  number  of  parts 
(eighths)  taken  or  considered,  thereby  obtaining  -f,  or  1,  as 
the  sum. 

(12)  When  the  denominators  are  not  alike,  we  know  that 
the  units  are  divided  into  unequal  parts,  so  before  adding 
them  we  must  find  a  common  denominator  for  the  denomi- 
nators of  all  the  fractions.  Reduce  the  fractions  to  fractions 
having  this  common  denominator,  add  the  numerators,  and 
write  the  sum  over  the  common  denominator. 

In  this  case,  the  least  common  denominator,  or  the  least 
number  that  will  contain  all  the  denominators,  is  16;  hence, 
we  must  reduce  all  these  fractions  to  16ths  and  then  add 
their  numerators. 

£  -(-  |  +  jV  —  '  To  reduce  the  fraction  %  to  a  fraction 
having  16  for  a  denominator,  we  must  multiply  both  terms 
of  the  fraction  by  some  number  which  will  make  the  denomi- 

1x4 
nator  16.     This  number  evidently  is  4,  hence,  -         =  T47. 

4:   /\    4 

Similarly,  both  terms  of  the  fraction  f  must  be  multiplied 

Q       \/     O 

by  2  to  make  the  denominator  16,  and  we  have  —          =  -^. 

O      X     & 

The  fractions  now  have  a  common  denominator  16;  hence, 
we  find  their  sum  by  adding  the  numerators  and  placing 
their  sum  over  the  common  denominator,  thus  :  T4¥  -f-  T6T  -f-  T8^ 


An, 


(13)  When  mixed  numbers  and  whole  numbers  are  to  be 
added,  add  the  fractional  parts  of  the  mixed  numbers  sep- 
arately, and  if  the  resulting  fraction  is  an  improper  fraction, 
reduce  it  to  a  whole  or  mixed  number.  Next,  add  all  the 
whole  numbers,  including  the  one  obtained  from  the  addi- 
tion of  the  fractional  parts,  and  annex  to  their  sum  the 
fraction  of  the  mixed  number  obtained  from  reducing  the 
improper  fraction. 


18  ARITHMETIC.  §  1 

42  +  31f  +  9T\  =  ?     Reducing  £  to  a  fraction  having  a 

5x2 

denominator  of  16,  we  have  —          =  -fg-.     Adding  the  two 

o  X  /* 

fractional    parts  of   the   mixed    numbers,    we   have  ^-g-  +  W 

_io^M_11_11 

16 

The  problem  now  becomes  42  +  31  +  9  +  1TV  —  ? 

Adding  all  the  whole  numbers  and  the  number  obtained 
from  adding  the  fractional  parts  of  the  mixed  numbers,  we 
obtain  83^  as  their  sum. 

42 
3  1 


8  3TV  Ans. 
(14)     291+501  +  41  +  69^=?     | 


_10       18,10,3   _ 

~~  TT-       tf  "T  Ttr  "T  IT  = 


The  problem  now  becomes  29  +  50  +  41  +  69  +  1TV  =  ? 

2  9      square  inches. 

5  0      square  inches. 
4  1      square  inches. 

6  9      square  inches. 
1T9T  square  inches. 

1  9  0^  square  inches.     Ans. 

(15)  |-  —  T7-g-  =  ?  When  the  denominators  of  fractions  are 
not  alike,  it  is  evident  that  the  units  are  divided  into  unequal 
parts  ;  therefore,  before  subtracting,  reduce  the  fractions  to 
fractions  having  a  common  denominator.  Then,  subtract 
the  numerators  and  place  the  remainder  over  the  common 
denominator. 


-=*.     An, 


§  1  ARITHMETIC.  19 

13  —  7TV  =  ?     This  problem  may  be  solved  in  two  ways: 

First  :    13  =  12|-|,  since  -ff-  =  1,  and  12|f  =  12  -f  -J-f  = 
12  +  1  =  13. 

We   can   now  subtract  the   whole   numbers   sepa- 
rately  and  the  fractions  separately,  and  obtain  12  —  7 

-j  n    _    iy 

«•    Ans. 


Second  '  :   By  reducing  both  numbers  to  improper  fractions 
having  a  denominator  of  16. 

1g_l3xl6_ao,  _(7xl6)  +  7      112  +  7      119 

=  T  -  T  X  16  -  ^  '  7™  -      ~W~  ~T6~  =  W- 

ft/\Q     _     1   1  Q 

Subtracting,    we   have   -8T°/  -  W  =         16         =  f|,  and 

H  =  16)89(5T97     the  same  result  that  was  obtained  by  the 
80  first  method. 

~^  312^  —  229^  =  ?      We  first  reduce  the 

—  fractions  of  the  two  mixed  numbers  to  frac- 

"I  r 

tions  having  a  common  denominator.    Doing 

9x2 

this,  we  have  ^  =  —          =  |f  .     We  can  now  subtract  the 
ID  X  &  , 

whole  numbers  and  fractions  separately,  and  have  312  —  229 
=  83  and       -       =         "-  =  83  +       =  83.     Ans. 


(16)  In  division  of  fractions,  invert  the  divisor  (or,  in 
other  words,  turn  it  upside  down)  and  proceed  as  in  multi- 
plication. 

-X-V  =  3[YT-1F-H^     Ans. 
)     A-3  =  iV-*-l=AXi  =  ^~  =  A  =  1V.     Ans. 

Y--9  =  -y--f  =  V-x-!  =  ^^  =  H.    Ans, 

H.     8.     I.— 27 


20  ARITHMETIC.  §  1 

=  4.     Ans. 


448)1808(4^ 
1792 


1  6    _  J_ 

448~28 

(e)     15f  -T-  4£  =  ?     Before   proceeding  with  the   division, 
reduce  both  mixed  numbers  to  improper  fractions.    Thus,  lof 


=  -3ff5-.     The  problem  is  now  -6T3-  -i-  -3-/-  =  ?     As  before,  invert 

9       2 

the  divisor,  multiply,  and  cancel  ;  -^  -T-  *-£-  =  -^  X  -/-$  =  j-  —  H? 

5 
=  J/  =  3|.     Ans. 

(17)  |.  rvalue  of   the  fraction,   and  28  the  numerator. 
We  find  that  4  multiplied  by  7  =  28,  so  multiplying  8,  the 
denominator  of  the  fraction,  by  4,  we  have  32  for  the  required 
denominator,  and  f  f  =  £.      Hence,  32  is  the  required  denom- 
inator. 

(18)  Since    these    four   bolts   measure    2£,   6|,    3TV,  and 
4  inches,  respectively,  together  they  will  measure  IG^-g-  inches, 
since  2|  +  6£  +  3TV  +  4  =  16TV       Reducing  the  fractions  of 
the  mixed  numbers  to  a  common  denominator,    we   have 


If  TV  of  an  inch  is 
allowed  for  cutting  and  finishing  each  bolt,  then  the  allow- 
ance for  the  4  bolts  would  equal  4  X  T\  =  ||  =  l|f  inches, 
-j^g  T  which  added  to  16^  inches  equals  18^-  inches, 

-p  2  the  length  of  the  piece  of  iron  required.     Ans. 


since       - 


ARITHMETIC. 

(PART  3.) 


(1)  A  fraction  is  one  or  more  of  the  equal  parts  of  a 
unit  and  is  expressed  by  a  numerator  and  a  denominator, 
while  a  decimal  fraction  is  a  number  of  tenths,  hundredths, 
thousandths,  etc.  of  a  unit  and  is  expressed  by  placing  a 
period  (.),  called  a  decimal  point,  to  the  left  of  the  figures 
of  the  numerator  and  omitting  the  denominator. 

(2)  To  reduce  the  fraction  £  to  a  decimal,  we  annex  one 
cipher  to  the  numerator,  which  makes  it  1.0.     Dividing  1.0, 
the  numerator,   by  2,   the  denominator,   gives   a   quotient 
of  .5,  the  decimal  point  being  placed  before  the  one  figure 
of  the  quotient,  or  .5,  since  only  one  cipher  was  annexed  to 
the  numerator. 


5.0  0000(.  15625 

3  2 

180 

160 


200 
192 


80 
64 
160 
160 

To  express  y6^  as  a  decimal,  write  the  numerator  and, 
beginning  with  the  right-hand  figure,  point  off  towards  the 
left  as  many  decimal  places  as  there  are  ciphers  in  the 

§X 

For  notice  of  copyright,  see  page  immediately  following  the  title  page. 


22  ARITHMETIC.  §  1 

denominator,  prefixing  ciphers  to  the  numerator  if  neces- 
sary, and  then  insert  the  decimal  point.  Proceeding  thus, 
jfo  =  • 65  and  w&  =  •  135.  Similarly,  T7rf£w  =  .  00024. 


Eight  hundredths. 


131  =  One  hundred  thirty-one  thousandths. 

£ 
ill 

«  a  3 

II    I 
III! 

0001  =  One  ten-thousandths. 


5^5  s5  a 

000027=  Twenty-seven  millionths. 


111 


.0108  =  One  hundred  eight  ten-  thousandths. 

nl 

If  11 

g  §  o  g 

9  3.0  1  0  1  =  Ninety-three  and  one  hundred  one  ten-  thousandths. 

In  reading  decimals,  read  the  number  just  as  you  would 

if  there  were  no  ciphers  before  it.     Then  count  from  the 


§  1  ARITHMETIC.  23 

decimal  point  towards  the  right,  beginning  with  tenths,  to 
as  many  places  as  there  are  figures,  and  the  name  of  the 
last  figure  must  be  annexed  to  the  previous  reading  of  the 
figures  to  give  the  decimal  reading.  Thus,  in  the  first 
example  given,  the  simple  reading  of  the  figure  is  eight  and 
the  name  of  its  position  in  the  decimal  scale  is  hundredths, 
so  that  the  decimal  reading  is  eight  hundredths.  Simi- 
larly, the  figures  in  the  fourth  example  are  ordinarily  read 
twenty-seven;  the  name  of  the  position  of  the  figure  7  in 
the  decimal  scale  is  millionths,  giving,  therefore,  the  decimal 
reading  as  twenty-seven  millionths. 

If  there  should  be  a  whole  number  before  the  decimal 
point,  read  it  as  you  would  read  any  whole  number  and 
read  the  decimal  as  you  would  if  the  whole  number  were 
not  there;  or,  read  the  whole  number  and  then  say  "and" 
so  many  hundredths,  thousandths,  or  whatever  it  may  be, 
as  "  ninety-three  and  one  hundred  one  ten-thousandths." 

(4)  In    addition    of    decimals,    the 
decimal  points  must  be  placed  directly  .125 
under  each  other,  so  that  tenths  will  •? 
come  under  tenths,  hundredths  under  .089 
hundredths,  thousandths  under  thou-  .4005 
sandths,    etc.     The    addition    is   then  -9 
performed  as  in  whole  numbers,  the  .000027 
decimal  point  of  the  sum  being  placed  2.2  1  4  5  2  7       Ans. 
directly    under    the    decimal    points 

above. 

(5)  A 


JiliJ 

Illlll 

.017 

.2 

.000047 


.217047=  T^vo  hundred  ^and  seventeen  thousand 
and  forty-seven  millionths.     Ans. 


24  ARITHMETIC.  §  1 

(6)     (a)   In    subtraction  of    decimals,    place  the  decimal 

7096300  P°intS    directly    under   each    other 

and  proceed  as  in  the  subtraction  of 

! whole  numbers,  placing  the  decimal 

7  0  8.7  7  8  6    Ans.    point    in    the    remainder    directly 
under  the  de'cimal  points  above. 

In  the  above  example,  we  proceed  as  follows:  We  cannot 
subtract  4  ten-thousandths  from  0  ten-thousandths,  and  as 
there  are  no  thousandths,  we  take  1  hundredth  from  the 
3  hundredths.  1  hundredth  =  10  thousandths  =  100  ten- 
thousandths.  4  ten-thousandths  from  100  ten-thousandths 
leaves  96  ten-thousandths.  96  ten-thousandths  =  9  thou- 
sandths -|-  6  ten-thousandths.  Write  the  6  ten-thousandths 
in  the  ten-thousandths  place  in  the  remainder.  The  next 
figure  in  the  subtrahend  is  1  thousandth.  This  must  be 
subtracted  from  the  9  thousandths  which  is  a  part  of  the 
1  hundredth  taken  previously  from  the  3  hundredths.  Sub- 
tracting, we  have  1  thousandth  from  9  thousandths  leaves 
8  thousandths,  the  8  being  written  in  its  place  in  the  remain- 
der. Next  we  have  to  subtract  5  hundredths  from  2  hun- 
dredths (1  hundredth  having  been  taken  from  the  3  hun- 
dredths leaves  but  2  hundredths  now).  Since  we  cannot  do 
this,  we  take  1  tenth  from  6  tenths.  1  tenth  =  10  hun- 
dredths, and  10  hundredths  +  2  hundredths  —  12  hundredths. 
5  hundredths  from  12  hundredths  leaves  7  hundredths. 
Write  the  7  in  the  hundredths  place  in  the  remainder.  Next 
we  have  to  subtract  8  tenths  from  5  tenths  (5  tenths  now, 
because  1  tenth  was  taken  from  the  6  tenths).  Since  this 
cannot  be  done,  we  take  1  unit  from  the  9  units.  1  unit  = 
10  tenths.  10  tenths  +  5  tenths  =  15  tenths,  and  8  tenths 
from  15  tenths  leaves  7  tenths.  Write  the  7  in  the  tenths 
place  in  the  remainder.  In  the  minuend  we  now  have 
708  units  (1  unit  having  been  taken  away)  and  0  units  in  the 
subtrahend.  0  units  from  708  units  leaves  708  units;  hence, 
we  write  708  in  the  remainder. 

(b)     8  1.9  6  3  (c)     1  8.0  0  (d)     1.000 

1.7  0  0  .18  .001 


8  0.2  6  3     Ans.          1  7.8  2     Ans.  .999     Ans. 


§  1  ARITHMETIC.  25 

(e)     872.1  —  (.8721  +  .008)  =  ?     In  this  problem  we  are  to 
subtract    (.8721 +  .008)    from    872.1.      First 
perform    the    operation  as  indicated   by  the       Q  0  8  0 
sign  between  the  decimals  enclosed  by   the 
parenthesis.  -8  8  °  l  sum- 

Subtracting  the   sum    (obtained  by   adding  the  decimals 
8721000  enclosed  within  the  parenthesis)  from 

8801  the  number  872.1  (as  required  by  the 

minus  sign   before   the  parenthesis), 


8  7  1.2  1  9  9     Ans.      we  obtain  the  required  remainder. 

(/)      (5.028  +  .0073)  —  (6.704  —  2.38)  =  ?     First  perform 
the  operations  as  indicated  by  the  signs 
between    the    numbers    enclosed    by    the 

parentheses.     The  first  parenthesis  shows         _1 

that  5. 028  and  .  0073  are  to  be  added.     This          5.0353  sum. 
Q  ij,  Q  ^  gives  5.0353  as  their  sum. 

2  3  g  Q  The  second  parenthesis  shows  that 

2. 38  is  to  be  subtracted  from  6. 704. 


4.3  2  4  diff.  The  difference  is  found  to  be  4  334. 

The    sign   between    the    parentheses   indicates    that     the 
quantities    obtained    by  performing  50353 

the  above  operations  are  to  be  sub-  43240 

tracted — namely,  that  4.324  is  to  be 

subtracted  from  5. 0353.      Performing  .7113    Ans. 

this  operation,  we  obtain  .7113  as  the  final  result. 

(7)  If  the  cost  of  the  coal  consumed  by  a  nest  of  steam 
.  boilers  amounts  to  $15.83  on  Monday,  to 

$14.70  on  Tuesday,  to  $14.28  on  Wednes- 

4  ^  g  day,  to  $13.87  on  Thursday,  to  $14.98  on 

Friday,  and  to  $12.65  on  Saturday,  then 

we  find  the  total  cost  of  the  week's  supply 

by  adding  the  different  amounts  together; 

hence,  $15.83  +  $14.70  +  $14.28  +  $13.87 

$86.31     Ans.      +  $14.98  +  $12.65  =  $86.31. 

(8)  482|  +  316£  +  390f  =  what  ? 


26  ARITHMETIC.  §  1 

When  mixed  numbers  are  to  be  added,  add  the  fractional 
parts  of  the  mixed  numbers  separately,  and  if  the  resulting 
fraction  is  an  improper  fraction,  reduce  it  to  a  whole  or 
mixed  number.  Next,  add  all  the  whole  numbers,  inclu- 
ding the  one  obtained  from  the  addition  of  the  fractional 
parts,  and  annex  to  their  sum  the  fraction  of  the  mixed 
number  obtained  from  reducing  the  improper  fraction. 
First,  we  will  reduce  the  fractional  parts  -f,  ^,  and  %  to 
equivalent  fractions  having  the  least  common  denominator. 
In  this  case  the  least  common  denominator  equals  the 
product  of  the  denominators  5,  3,  and  4,  since  we  cannot 
divide  any  two  of  them  by  any  number  (except  1)  without 
having  a  remainder,  as  can  be  done  in  the  examples  in 
Art.  26,  Part  2.  Hence,  the  least  common  denominator 
=  5  X  3  X  4  =  60.  Reducing  f  ,  £,  and  f  to  fractions  having 
this  least  common  denominator,  we  have  60  divided  by 
the  first  denominator,  5,  equals  12.  Then,  \  x  \\  =  •£$  . 
GO  divided  by  the  second  denominator,  3,  equals  20.  Then, 
$  X  f  £  =  f-jr-  60  divided  by  the  third  denominator,  4,  equals 
15.  Then,  f  X  TT  —  TO-  The  sum  °f  these  fractions  equals 


The  problem  now  becomes  482  +  316  +  390  -f  Iff,  the  sum 
of  which  equals 


482 

316 

390 

Iff 

1  189-ff 

1500        represents  the  actual  horsepower  required. 

1  1  8  9  -f  f  represents  the  indicated  horsepower  of  the  engines 


310  TV>  or  310. llf  =  the   H.    P.   to  be  developed  by  the 
new  engine.     Ans. 


§  1  ARITHMETIC.  27 


7 
reduced  to  its  equivalent  decimal  = 


6  0 


100 
6  0 


(9)  Since    the    inside    diameter    of  the    steam     pipe    is 
C.06  inches  and  the  outside   diameter  is  6.62  inches,  there 
is  a  difference  of  6.62  —  6.06,  or  .56  of  an  inch,  in  both  diam- 
eters.   But  .56  of  an  inch  is  just  twice  the  thickness  of  the 
pipe;  hence,  the  pipe  is  £  of  .56,  or  .28  of  an  inch  thick. 

(10)  (a)  There  are   3  decimal  places  in  the  multiplicand 
.107  and  3  in  the  multiplier;  hence,  there 
.013                      are  3  +  3,  or  6,  decimal  places  in  the 

321  product.      Since    the    product    con- 

107  tains  but  four  figures,  we  prefix  two 

ciphers  in  order  to  obtain  the  neces- 


sary 6  decimal  places. 

(&}      203  There  are  2  decimal  places  in  the 

^•03  multiplier  and  none   in  the  multipli- 

509  cand;  hence,    there  are    2  +  0,  or  2, 

000  decimal  places  in  the  first  product. 

406  Since  in  the  second  multiplication 

there  are  2  decimal  places  in  the  mul- 
tiplicand and  3  decimal  places  in  the 
multiplier,  there  are  3  +  2,  or  5,  deci- 
123627  mal  places  in  the  second  product. 

00000  When  there  are  one  or  more  ciphers 

82418  in  the  multiplier,   multiply  just  the 

83.65427     Ans.          same  as  with  the  other  figures. 

(c)     First  perform  the  operations  indicated  by  the  signs 

between    the   numbers   enclosed    by  the    parentheses,    and 

then  whatever  may  be  required  by  the  sign  between  the 
parentheses. 


28  ARITHMETIC.  §  1 

3  1.8  5 

The  first  parenthesis    shows  that 


22295  the  numbers  2.7  and  31.85  are  to  be 

6370  multiplied  together. 

8  5.9  9  5 

The  second  parenthesis  shows  that 
.316  is  to  be  taken  from  3.16. 


2.8  4  4 

The  product  obtained  by  performing  the  operation  indi- 

8  5.9  9  5  cated    by    the   signs    within  the    first 

2.8  4  4  parenthesis  is  now  multiplied  by  the 

343980  remainder  obtained  by  performing  the 

343980  operation  indicated  by  the  signs  within 

687960  t^ie  second  parenthesis. 
71990 


2  4  4.5  69780     Ans. 

(d)     (107.8  +  6.541-31.96)  X  1.742=  ? 

I  0  7.8  8  2.3  8  1 
+         6.541                         X      1.742 

II  4.3  41  164762 
-      31.96                          329524 

8  2.3  8  1  576667 

82381 


1  4  3.5  0  7  7  0  2     Ans. 

(11)     If  one  3-inch  tube  measures  15£  ft.  in  length,  60  of 

these    tubes    would    measure    60  X  15£   ft.,    or 

15^  930  ft.  in  length.     If  1  foot  of  tubing  heats  a 

6  0  surface  of  .728  sq.  ft.,  then  it  is  evident  that 

900  930  ft.  of  tubing  would  hea't  a  surface  of  930 

30  X  .728  ft.,  or  677.04  sq.  ft. 

930  ft.  .728 

930 
21840 
6552 


6  7  7.0  4  0  sq.  ft. 


§  1  ARITHMETIC.  29 

(12)     (a)  l  =  7-^tV=7xY  =  ^-^  =  HjL  =  374. 

™  Ans. 

The  heavy  line  indicates  that  7  is  to  be  divided  by  TV 

<"     1=1  +  1  =  1*!  =     =  ^     An, 

"8"  4 


, 


1.25  X  20  X  3  _  ,       In    this    problem    1.25  X  20  X  3 
87  -j-  88  "  constitutes  the   numerator  of   the 


459  -f-  32  complex  fraction. 

1.2  5 

X  20 

-  -          Multiplying  the  factors  of  the  numerator 

together,  we  find  their  product  to  be  75. 
X          o 

&*y  I  CG 
The    fraction    -rz-—       -r   constitutes    the    denominator    of 


-r 

O/i 

the  complex  fraction.     The  value  of  the  numerator  of  this 
fraction  equals  87  +  88  =  175. 

The  value  of  the  denominator  of  this  fraction  is  equal  to 
459  -f-  32  =  491.  The  problem  then  becomes 

75  3 

75___1__75_^175_75      491     ??  x  491     1,473  3 

175      175       1    '  491       1       175          ;jj5  7  f 

491      491  7  Ans. 

(13)  The  pitch  of  the  rivets  is  the  distance  between  the 
centers  of  the  rivets.  Hence,  since  the  distance  around  the 
cylindrical  boiler  is  166.85  in.,  and  there  are  72  rivets  in  one 
of  the  seams,  the  pitch  of  the  rivets  equals  166.85-7-72 
=  2.  317+  in.  Ans. 

7  2)  1  6  6.8  5  0  (2.3  1  7+ 
144 


30  ARITHMETIC.  §  1 

(14)  If  a  keg  containing  133  boiler  rivets  weighs 
100  pounds,  then  each  rivet  must  weigh  as  much  as  133  is 
contained  times  in  100,  or  .  75  of  a  pound. 

1  3  3)  1  0  0.00  (.7  5+     Ans. 
931 


690 
665 

2  5 


Since  there  are  2  decimal  places  in  the  dividend  and 
0  decimal  places  in  the  divisor,  we  must  point  off  2  —  0 
=  2  decimal  places  in  the  quotient,  or  answer. 


1  foot  =  12  inches. 

3 

I-  of  1  foot  =  \  X  ~  =  ~  =  10i  inches.     Ans. 
p         1          A 

2 
(16)     12  inches  =  1  foot. 

Q  -J 

T\  of  an  inch  =  T3¥  -4-  12  =  ^  X  ^  =  -fa  of  a  foot. 

4 

J_ 

6~4  )  1.0  0  0  0  0  0  (.0  1  5  6  2  5     Ans. 
6  4 

360 

320 

.  Point  off  6  decimal  places  in 

„  the  quotient,  since  we  annexed 

six  ciphers  to  the  dividend,  the 
1^0  divisor   containing   no  decimal 

places;  hence,  6  —  0  —  6  places 
320  to  be  pointed  off. 

320 


§  1  ARITHMETIC.  31 

(17)  The  total  horsepower  developed  equals  48.63 
+  45. 7  +  46. 32  +  47. 9  +  48. 74  +  48. 38  +  48. 59  =  334. 26. 

Since  the  horsepower  developed  equals  334.26,  then  the 
average  horsepower  developed  must  equal  334.26  -f-  7,  or 
47.75+H.  P. 

48.6  3 
45.7 

4  6.3  2 
4  7.9 

48.7  4 
48.3  8 
4  8.5  9 


7  )  3  3  4.2  6 

47.75+     Ans. 


ARITHMETIC. 

(PART  4.) 


(1)  A  certain    per  cent,    of    a    number   means  so  many 
htmdredths  of  that  number. 

25$  of   8,428  Ib.  means  25  hundredths  of  8,428  Ib.     -^ 
=  .25.     Hence,  8,428  Ib.  X  .25  =  2,107  Ib.     Ans. 

(2)  \%  means  one-half  of  1  per  cent.     Since  \%  is  .01,  \<f> 
is  .005,  for  2)  'HI-     And  $35,000  X  .005  =  1175.     Ans. 

$  3  5  0  0  0 
.005 


$  1  7  5.0  0  0 

(3)  If  2  is  a  certain  per  cent,  of  50,  then  50  multiplied 
by  a  certain  rate  gives  a  product  of  2,  and  that  rate  is  equal 
to  2  divided  by  50.      Dividing  2  by  50, 

the  quotient  is  .04,  which  means  that    5  0  )  2.0  0  ( .0  4   Ans. 
2   is   4$   of   50,   or,  since   percentage 
=  base  X  rate, 

rate  =  percentage  -r-  base 

=  2-^  50  =  .04,  or  4#.     Ans. 

(4)  Since  percentage  =  base  X  rate,    rate  =  percentage 
-T-  base. 

As    percentage  =  10    and  base  =  10,    we    have    rate  =  10 
-f-  10  =  1. 

§2 

For  notice  of  copyright,  see  page  immediately  following  the  title  page. 


ARITHMETIC.  §2 


But  l  —  i^-o  and  f££  =  100$;  hence,  the  rate  (1)  means 
that  10  is  100$  of  10. 

(5)  Since  5,500  Ib.  represent  an  increase  of  15$  over  the 
consumption  when  the  condenser  is  used,  5,500  Ib.  must  be 
the  amount,  .15  the  rate,  and  the  number  of  pounds  con- 
sumed when  the  condenser  is  running  (to  be  found)  the 
base. 

Base  =  amount  -^  (1  +  rate)  =  5,500  -h  (1  +  .15) 
=  5,500  -h  1.15  -  4,782.61  Ib.,  nearly.     Ans. 

1.1  5  )  5  5  0  0.0  0  0  0  (  4  7  8  2.6  1 
460 


0  0 
0  5 


950 
920 


300 
230 


700 
690 


100 
115 

Or,  this  problem  could  also  have  been  solved  as  follows: 
100$  =  the  number  of  pounds  consumed  when  the  con- 
denser is  running.  If  there  is  a  gain  of  15$,  then  100$  +  15$, 
or  115$  =  5,500  Ib.,  the  amount  used  when  the  condenser 
is  not  running.  If  115$  =  5,500  Ib.,  1$  =  Tfg-  of  5,500 
=  47.8261  Ib.,  and  100$  =  100  X  47.8261  =  4,782.61  Ib.  Ans. 

(6)  24  $  of  $950  =  950  X  .24    =  $228 

12£$of  $950  =  950  X  .125=    118.75 
17  $  of  $950=  950  X  .17    =    161.50 

53£$  of  $950  =  $508.25 

The  total  amount  of  his  yearly  expenses,  then,  is  $508.25; 
hence,  his  savings  are  $950  —  $508.25  =  $441.75.     Ans. 


2  ARITHMETIC.  3 


Or,  as  above,  24$  +  12£<£  +  17$  =  53|#,  the  total  percent- 
age of  expenditures;  hence, 

$950  X  .535=  $508.25,  and 

$950  —  $508.25  =  $441.75  =  his  yearly  savings.     Ans. 

(7)     The  percentage  is  961.38  and  the  rate  is  .37£. 

Base  =  percentage  -f-  rate 

=  9G1.38  -5-  .375  =  2,563.68,  the  number.     Ans. 

.3  7  5)  9  6  1.3  8  000  (2  5  6  3.6  8 
750 


2113 

Another  method  of  solv-  1875 

ing  is  the  following:  «  o  o  o 

If  37|$  of  a  number  is  2250 

961.38,  then  .37^  times  the 


number  =961. 38   and    the 

number  =  961.38  -=-   .37$,  l  l  2  5 


which,  as  above  =  2, 563. 68.  2550 

Ans.  2250 


3000 
3000 

(8)  298  revolutions  per  minute  with  the  load  =  base, 
.01£  =  rate,  and  the  amount  (to  be  found)  will  equal  the 
speed  of  the  engine  when  running  unloaded. 

Amount  =  base  X  (1  +  rate) 

—  298  X  (1  +  .015)  =  302.47  rev.  per  min.     Ans 

o  &  9  8 
X      1.0  1  5 


1490 
298 
000 
298 

3  0  2.4  7  0 

//.  8.  1.—38 


4  ARITHMETIC.  §  2 

(9)  4  yd.   2  ft.   10  in.  to  inches. 
X          3 

1  2  Since    there    are    3    feet    in 

+         2  1  yard,  in  4  yards  there  are  4x3 

^  £eet  feet,  or  12    feet.      12  feet    plus 

x       1  2  2  feet  =  14  feet- 

There  are  12  inches  in  1  foot ; 

therefore,  in  14  feet  there  are 
12  X  14,  or  168  inches.  168  inches 

168  plus  10  inches  =  178  inches. 

+  _1_0 

178  inches.     Ans. 

(10)  1  2  )  3  7  2  2  inches. 

3)31  0  +  2  inches. 
103  +  1  foot. 

Ans.  =  103  yd.  1  ft.  2  in. 

EXPLANATION. — There  are  12  inches  in  1  foot;  hence,  in 
3,722  inches  there  are  as  many  feet  as  12  is  contained  times 
in  3,722,  or  310  feet  and  2  inches  remaining.  Write  2  inches 
as  a  remainder.  There  are  3  feet  in  1  yard;  hence,  in 
310  yards  there  are  as  many  feet  as  3  is  contained  times 
in  310,  or  103  yards  and  1  foot  remaining.  Hence,  in 
3,722  inches  there  are  103  yd.  1  ft.  2  in. 

(11)  1728  )  7  6  4  3  2  5  cu.  in. 

27)442  +  549  cu.  in. 

1  6  cu.  yd.  +  10  cu.  ft. 

Ans.  =  16  cu.  yd.  10  cu.  ft.  549  cu.  in. 

EXPLANATION. — There  are  1,728  cubic  inches  in  1  cubic 
foot;  hence,  in  764,325  cu.  in.  there  are  as  many  cubic  feet 
as  1,728  is  contained  times  in  764,325,  or  442  cubic  feet 
^and  549  cubic  inches  remaining.  Write  the  549  cubic  inches 
as  a  remainder.  There  are  27  cubic  feet  in  1  cubic  yard; 
hence,  in  442  cubic  feet  there  are  as  many  cubic  yards  as  27 


§  2  ARITHMETIC.  5 

is  contained  times  in  442  cubic  feet,  or  16  cubic  yards  and 
10  cubic  feet  remaining.  Then,  in  704,325  cubic  inches 
there  are  16  cu.  yd.  10  cu.  ft.  549  cu.  in. 

(12)  T.     cwt.     Ib.  Since     in    1    ton    there    are 
16        8        7  5       20  cwt.,    in   16   tons  there    are 

X         2  0  16  X  20  =  320    cwt.      320    cwt. 

320  -f  8  cwt.  =  328  cwt.     There  are 

+  8  100    Ib.    in    1    cwt. ;    hence,    in 

328  cwt.  328    cwt.    there    are    328  X  100 

X  100  =  32,800  Ib.     32,800  lb.  + 75  lb. 

'32800  =  32,875  Ib.     Ans. 
+              75 

3  2  8  7  5  Ib.     Ans. 

(13)  100  )  2  5  3  9  6  Ib. 

2  0  )  2  5  3  cwt.  +  96  Ib. 
1  2  T.  -f  13  cwt. 

There  are  100  Ib.  in  1  cwt. ;  hence,  in  25,396  Ib.  there  are 
as  many  cwt.  as  100  is  contained  times  in  25,396,  or  253  cwt. 
and  96  Ib.  remaining. 

There  are  20  cwt.  in  1  ton,  and  in  253  cwt.  there  are  as 
many  tons  as  20  is  contained  times  in  253,  or  12  tons  and 
13  cwt.  remaining.  Hence,  25,396  -lb.  =  12  T.  13  cwt. 
96  lb.  Ans. 

(14)  Arrange  the  different  terms  in  columns,  taking  care 

to  have  like  denominations  in  the 

yd.      ft.       in.  same   column.     We  begin  to  add 

at  the  right-hand  column.      7  +  9 

419  +3  =  19  in.  ;  since  12  in.  =  1  ft., 

%         7  19  in.  =  1  ft.  and  7  in.     Place  the 

807      Ans.      7   in.    in  the   inches   column    and 

reserve  the  1  ft.  to  add  to  the  sum 

of  the  feet.  2  +  1  +  2  +  1  (reserved)  =  6  ft.  Since  3  ft. 
=  1  yd.,  6  ft.  =  2  yd.  and  0  ft.  remaining.  Place  the  0  in 
the  feet  column  and  reserve  the  2  yd.  to  add  to  the  sum  of 
the  yards.  4  +  2  -f-  2  (reserved)  =  8  yd.,  which  we  place  in 
yards  column.  Ans.  =  8  yd.  7  in. 


G  ARITHMETIC.  §  2 

(15)  Since  10  gal.  2  qt.  1  pt.  of  machine  oil  is  sold  at  one 

time  and  16  gal.  3  qt.  at  another 

gal.       qt.      pt.  time,     together    there     was    sold 

1021  27  gal.  1  qt.  1  pt.   0  +  1  =  1  pt.   We 

16         3         0  cannot  reduce  1  pt.  to  any  higher 

27         1         1  denomination,    so  place    it    under 

pints  column.   3  qt.  +2  qt.  =  5  qt. 

Since  4  qt.  =  1  gal.,   5   qt.  =  1  gal.   and   1  qt.   remaining. 

Place  1  qt.  under  quarts  column  and  reserve  the  1  gal.  to  add 

to  the  gallons.     16  gal.  -f  10  gal.  +  1  gal.  (reserved)  =  27  gal. 

Since  the  barrel  contained  31£,  or 

gal-        qt-      Pt-  31. 5  gal.,   and  27  gal.   1  qt.   1  pt. 

31         2         0  were  sold,  there  remained  the  dif- 

27         1         1  ference,  or  4  gal.  1  pt.     31.5  gal. 

4         0         1      Ans.      =  31  gal.   2  qt.,   since  .5  =  |,  and 

|  of  1   gal.  =  %  of  4   qt.  =  2   qt. 

1  pt.  cannot  be  taken  from  0  pt.,  so  we  take  1  qt.  from 
the  2  qt.  The  1  qt.  taken  =  2  pt.  1  pt.  from  2  pt.  =  1  pt. 
Place  1  pt.  under  pints  column.  Since  we  took  1  qt.  from 
the  quarts  column,  there  remains  2  —  1,  or  1  qt.  1  qt.  from 
1  qt.  leaves  0  qt.  Place  0  qt.  under  the  quarts  column. 
27  gal.  from  31  gal.  leaves  4  gal.  Place  4  gal.  under  the 
gallons  column.  We  therefore  find  that  4  gal.  1  pt.  of 
machine  oil  remained  in  the  barrel. 

(16)  In  multiplication  of  denominate  numbers,  we  place 
the  multiplier  under  the  lowest  denomination  of  the  multi- 
plicand, as 

1  7  ft.     3  in. 
5  1 


8  7  9  ft.    9  in. 

and  begin  at  the  right  to  multiply.  51x3  =  153  in.  Since 
there  are  12  in.  in  1  ft.,  in  153  in.  there  are  as  many  feet  as 
12  is  contained  times  in  153,  or  12  ft.  and  9  in.  remaining. 
Place  the  9  in.  under  the  inches  and  reserve  the  12  ft.  51  X  17 
ft.  =  867  ft. ;  867  ft.  +  12  ft.  (reserved)  =  879  ft.  879  ft. 


§  2  ARITHMETIC.  7 

can  be  reduced  to  higher  denominations  by  dividing  by  3  ft. 
to  find  the  number  of  yards,  and  by  5£  to  find  the  number 
of  rods. 

3  )  8  7  9  ft.  9  in. 
5.5  )  2  9  3  yd. 

5  3  rd.  1£  yd. 

Then,  879  ft.  9  in.  =  53  rd.  Ifc  yd.  0  ft.  9  in.,  or  53  rd.  1  yd. 
2  ft.  3  in. 

(17)  Since  2  pt.  =  1    qt.,    3   qt.  =  3  X  2,  or  6  pt.     6  pt. 
+  lpt.  =  7  pt.     4.7  X  7=  32.9  pt.     Ans. 

qt.       pt.  7  pt. 

3         1  4.7 

X_2  —9 

6  28 

+_Jl  3~2~9  pt. 

7Pt. 

(18)  If  there  are  four  lengths,  each  15  ft.  5   in.,  15   ft. 
5  in.  X  4  =  60  ft.  20  in.,  or  the  length  of  the  four  pieces. 


14  8 


8         10  6  0  ft.     2  0  in. 


82         38,  or  85  ft.  2  in.  =  the  length  of  the  shaft. 

From  the  length  of  the  shaft  we  must  subtract  8  in.  X  2 
=  16  in.  to  get  the  distance  between  the  end  hangers. 

ft.          in. 

82         38 

1  6 


82         22,  or  83  ft.  10  in. 

Since  there  are  six  hangers,  there  are  five  spaces.     The 
length  of  one  space  is  83  ft.  10  in  -4-  5  =  16  ft.  9-J-  in.     Ans. 


8  ARITHMETIC.  §  2 

(19)     Reducing  18  ft.  11£  in.  to  inches,  we  have  227£  in., 
or  227.25  in. 

ft.  in. 

18         Hi 
1  2 

3  6 
1  8 


216 
11* 

2  2  7  i  in. 

1£  X  2,  or  2^  in.,  for  the  two  end  rivets. is  deducted  from 
the  length,  leaving  225  in.,  which  is  divided  into  equal 
spaces  by  the  rivets. 

2  2  7.2  5  in. 
-      2.2  5  in. 


225         in. 

The  pitch  of  the  rivets   (or  the  distance  between  their 
centers)  is  1^  in.,  or  1.25  in. ;  hence, 

1.2  5  )  2  2  5.0  0  (  1  8  0     225  -h  1.25  =  180  spaces  between  the 
125  rivets.     But  since  there  will  be  one 

..  Q  Q  Q  more    rivet    than    the    number    of 

1000  spaces,   the    number    of    rivets    re- 

quired for  this  boiler  shell  will  be 
180  +  1  —  181.     Ans. 


ARITHMETIC. 

(PART  5.) 


(1)  To  find  the  second  power  of  a  number,  we  must  mul- 
tiply the  number  by  itself  once;  that  is,  use  the  number 
twice  as  a  factor.  Thus,  the  second  power  of  108  is  108 
X  108  =  11,664. 

108 

108 


1  81.25 
1  81.25 


9  06  25 
36250 
18125 
145  000 
181  25 

328  51.56  25 
1  81.25 

1642578125 

6570  31250 
32851 5625 
2628125  000 
328515625 

595434  5.7  031  2  5  Ans. 


(2)  The  third  power  of  181. 25 
equals  the  number  obtained  by 
using  181.25  as  a  factor  three 
times.  Thus,  the  third  power 
of  181.25  is  181.25  X  181.25 
X  181.25  =  5,954,345.703125. 

Since  there  are  2  decimal 
places  in  the  multiplier  and  2 
in  the  multiplicand,  there  are 
2  +  2  =  4  decimal  places  in  the 
first  product. 

Since  there  are  4  decimal 
places  in  the  multiplicand  and 
2  in  the  multiplier,  there  are 
4  +  2  =  6  decimal  places  in  th<i 
final  product. 
§2 


For  notice  of  copyright,  see  page  immediately  following  the  title  page. 


10  ARITHMETIC.  §  2 

(3)     (a)  .01333  =  .0133  X  .0133  X  .0133  =  .000002352637. 
.0133  Since  there  are  4  decimal 

0  i  3  3  places    in    the    multiplicand 

—  and  4  in  the   multiplier,  we 

j*  should    point    off    4  +  4  =  8 

decimal  places  in  the  prod- 
uct;   but    as  there    are  only 


.00017689  5  figures  in  the  product,  we 

•0133  prefix  3  ciphers  to  form  the 

53067  8    necessary    decimal    places 

53067  m  the  first  product. 

17689  Since  there  are  8  decimal 


.000002353637  Ans. 

and   4  in   the  multiplier,  we 

should  point  off  8  +  4  =  12  decimal  places  in  the  product; 
but  as  there  are  only  7  figures  in  the  product,  we-  prefix 
5  ciphers  to  make  the  necessary  12  decimal  places  in  the  final 
product. 

(b)     301.0111  =  301.011  X  301.011  X  301.011 

=  27,273,890.942264331. 
3  0  1.0  1  1 
3  0  1.0  1  1 


301011 
301011 
301011 
903033 

9060  7.6  22121 
3  0  1.0  1  1 

90607622121 
90607622121 
90607622121 
271822866363 

2727389  0.9  4  2  264331  Ans. 

Since  there  are  3  decimal  places  in  the  multiplicand  and 
3  in  the  multiplier,  we  should  point  off  3  +  3=6  decimal 
places  in  the  first  product. 


§  2  ARITHMETIC.  11 

Since  there  are  6  decimal  places  in  the  multiplicand  and 
3  in  the  multiplier,  we  should  point  off  6  +  3  —  9  decimal 
places  in  the  final  product. 

W     (W  =  *X*Xi  =  l£i£|  =  Th.     Ans.  xj 

6  4 

X      8 


512 


(d]  To  find  any  power  of  a  mixed  number,  first  reduce  it 
to  an  improper  fraction,  and  then  multiply  the  numerators 
together  for  the  numerator  of  the  answer  and  multiply  the 
denominators  together  for  the  denominator  of  the  answer. 


.  =  Y  X  -V  X  Y-  =  =     = 

Ans. 


15  64)3375.000000(52.734375 
15          320 

T5  175 

15  128 

225  47  0 

15  448 


1125  220 

225  192 


3375  280 

256 


240 
192 

480 
448 


320 
320 


ARITHMETIC. 


Since  6  ciphers  were  annexed  to  the  dividend,  0  decimal 
places  must  be  pointed  off  in  the  quotient. 


(a) 


(d)  20 


28 


|/3'4  8'6  7'8  4.4  O'l  0  =  1  8  6  7.2  9  + 
(6)  1 

(0  248 
224 

(e)   2467 
2196 


Ans. 


27184 
26089 

109540 

74684 

3485610 
3361041 

124569 

EXPLANATION. — (I)  Divide  the  number 
into  periods.  In  the  above  case,  where  the 
number  consists  of  a  whole  number  and  a 
decimal,  we  begin  at  the  decimal  point  and 
proceed  to  the  left  in  pointing  off  the 


3720 

7 

3727 
7 

37340 

2 

37342 

2 

373440    wJwle  number  and  towards  the   right   in 
9    pointing  off  the  decimal,  annexing  a  0  to 
complete  the  last  decimal  period.    In  square 
root  two  figures  constitute  a  period. 

Find  the  greatest  single  number  whose  square  is  less  than 
or  equal  to  3,  the  first  period.  This  is  evidently  1,  since 
2"  =  4,  which  is  greater  than  3.  Write  1  as  the  first  figure 
of  the  root;  also  write  it  to  the  left,  as  shown  at  (a). 
Now,  multiply  the  1  at  (a]  by  the  1  in  the  root  and  write 
the  result  under  the  first  period,  or  3,  as  shown  at  (b). 
Subtract,  bring  down  the  next  period,  48,  and  annex  it  to 
the  remainder  2,  thus  making  248  the  dividend,  as  shown 
at  (c).  Add  the  root  already  found  to  the  1  at  (<?),  thereby 


§  2  ARITHMETIC.  13 

obtaining  2,  and  annex  a  cipher  to  this  2,  thus  making  it  20, 
which  we  call  the  trial  divisor. 

Divide  the  dividend  248  at  (c)  by  the  trial  divisor  20  at 
(d]  and  obtain  8,  which  is  probably  the  next  figure  of  the 
root.  Write  8  in  the  root,  as  shown,  and  also  add  it  to  20, 
the  trial  divisor,  making  it  28.  This  is  called  the  complete 
divisor. 

(II)  Multiply  the  complete  divisor  28  by  8,  the  second 
figure  of  the  root,  and  subtract  the  result  from  the  divi- 
dend (c).     The  remainder  is  24,  to  which  annex  the  next 
period,  making  2,467,  as  shown  at  (e),  which  call  the  neiv 
dividend. 

Add  the  second  figure  of  the  root,  or  8,  to  the  divisor  28 
and  annex  a  cipher,  thus  obtaining  360.  Dividing  2467  by 
360,  we  find  6  to  be  the  next  figure  of  the  root.  Adding 
this  last  figure  of  the  root,  or  6,  to  360,  we  get  366,  and 
multiplying  by  this  last  figure  of  the  root,  or  6,  gives  2196, 
which  we  write  under  2467  and  subtract. 

(III)  Annexing  the  next  period,  84,  to  the  remainder  271 
gives  27184   as  the  next  new  dividend.      Now,   adding  the 
third  figure  of  the  root  to  366  and  annexing  a  cipher,  as 
before,  we  have  3720.     Dividing  27184  by  3720,  the  result 
is  7,  which  write  as  the  next  figure  of  the  root.     Adding  the 
fourth  figure  of  the  root,  or  7,  to  3720,  we  get  3727,  and 
multiplying  by  7  of  the  root  gives  26089,  which  write  under 
27184  and  subtract,  obtaining  1095  as  a  remainder. 

(IV)  Annexing  the  next  period,  or  40,  to    the    remain- 
der 1095  gives  109540  as  the  next  new  dividend.     Adding 
the  last  figure  of  the  root  to  3727  and  annexing  a  cipher, 
as  before,  the  result  is  37340.     Dividing  109540  by  37340, 
the  result  is  2,  which  write  as  the  next  figure  of  the  root. 
Adding  the  fifth  figure  of  the  root,  or  2,  to  37340,  we  obtain 
37342,  and  multiplying  by  2  of  the  root  gives  74684,  which 
write    under    109540    and    subtract,    obtaining    34856    as    a 
remainder. 

(V)  Annexing  the  next  period,  or  10,  to  the  remainder 
34856  gives  3485610  as  the  next  new  dividend.     Now,  adding 


14  ARITHMETIC.  §2 

the  last  figure  of  the  root  to  37342  and  annexing  a  cipher, 
as  before,  the  result  is  373440.  Dividing  3485610  by  373440 
gives  9  as  a  result,  which  write  as  the  next,  or  last  figure, 
of  the  root.  Adding  the  last  figure  of  the  root,  or  9,  to 
373440,  we  get  373449,  and  multiplying  by  9  of  the  root 
gives  3361041,*  which  write  under  3485610  and  subtract. 
Since  there  is  a  remainder,  we  know  the  given  power  is  not 
a  perfect  square,  so  we  place  -j-  after  the  root. 

In  this  problem  there  are  six  periods — four  in  the  whole 
number  and  two  in  the  decimal — hence,  there  will  be  six 
figures  in  the  root  (since  we  obtain  one  figure  of  the  root  for 
each  period),  four  figures  constituting  the  whole  number  and 
two  figures  the  decimal  of  the  root.  Hence, 

4/3,486,784.401  =  1,867.29+. 


(a) 
(d) 

3           4/9'0  0'00'9  9.4  0*0  9'0  0  =  3000.01  6  + 
3       (b)    9 

60   (c) 
0 

6~00 
0 

0000994009 
600001 

39400800 
36  000156 

6000 
0 

3400644 
EXPLANATION.  —  Beginning  at  1 

60000 
0 

600000       decimal  point,  we  point  off  the  whole 
1       number  into  periods  of  two  figures 
each,  proceeding  from  right  to  left; 
also    point    off    the    decimals   into 

periods  of  two  figures  each,  proceed- 

6000020    ing  from  left  to  right.     The  largest 

6    number  whose  square  is  contained 

6000026    m  the  first  period,  9,  is  3 ;  hence,  3 

is  the  first  figure  of  the  root.     Place 

3  at  the  left,  as  shown  at   (#),  and  multiply  it  by  the  first 
figure  in  the  root,  or  3.     The  result  is  9.    Write  9  under  the 


§  2  ARITHMETIC.  15 

first  period,  9,  as  at  (b),  subtract,  and  there  is  no  remainder. 
Bring  down  the  next  period,  which  is  00,  as  shown  at  (c). 
Add  the  root  already  found  to  the  3  at  (a),  obtaining  6,  and 
annex  a  cipher  to  this  6,  thus  making  it  60,  which  is  the  trial 
divisor,  as  shown  at  (</).  Divide  the  dividend  (c]  by  the 
trial  divisor  and  obtain  0  as  the  next  figure  in  the  root. 
Write  0  in  the  root  as  shown,  and  also  add  it  to  the  trial 
divisor  60,  and  annex  a  cipher,  thereby  making  the  next  trial 
divisor  600.  Bring  down  the  next  period,  00,  annex  it  to  the 
dividend  already  obtained,  and  divide  it  by  the  trial  divisor. 
600  is  contained  in  0000  no  times,  so  we  place  another  cipher 
in  the  root.  Write  0  in  the  root,  as  shown,  and  also  add  it 
to  the  trial  divisor  600,  and  annex  a  cipher,  thereby  making 
the  next  trial  divisor  6000.  Bring  down  the  next  period,  99. 
The  trial  divisor  6000  is  contained  in  000099  no  times,  so 
we  place  0  as  the  next  figure  in  the  root,  as  shown,  and  also 
add  it  to  the  trial  divisor  6000  and  annex  a  cipher,  thereby 
making  the  next  trial  divisor  60000.  Bring  down  the  next 
period,  40,  and  annex  it  to  the  dividend  already  obtained 
to  form  the  new  dividend,  00009940,  and  divide  it  by  the 
trial  divisor  60000.  60000  is  contained  in  00009940  no  times, 
so  we  place  another  cipher  in  the  root,  as  shown,  and 
also  add  it  to  the  trial  divisor  60000  and  annex  one  cipher, 
thereby  making  the  next  trial  divisor  600000.  Bring  down  the 
next  period,  09,  and  annex  it  to  the  dividend  already  obtained 
to  form  the  new  dividend,  0000994009,  and  divide  it  by  the 
trial  divisor  600000.  600000  is  contained  in  0000994009  once, 
so  we  place  1  as  the  next  figure  in  the  root  and  also  add 
it  to  the  trial  divisor  600000,  thereby  making  the  complete 
divisor  600001.  Multiply  the  complete  divisor  600001  by  1, 
the  sixth  figure  in  the  root,  and  subtract  the  result  obtained 
from  the  dividend.  The  remainder  is  394008,  to  which  we 
annex  the  next  period,  00,  to  form  the  next  new  dividend, 
or  39400800.  Add  the  sixth  figure  of  the  root,  or  1,  to  the 
divisor  600001  and  annex  a  cipher,  thus  obtaining  6000020 
as  the  next  trial  divisor.  Dividing  39400800  by  6000020, 
we  find  6  to  be  the  next  figure  of  the  root.  Adding  this  last 
figure,  6,  to  the  trial  divisor,  we  obtain  6000026  for  our  next 


16  ARITHMETIC.  §  2 

complete  divisor,  and  multiplying  by  the  last  figure  of  the 
root,  or  G,  gives  36000156,  which  write  under  39400800  and 
subtract.  Since  there  is  a  remainder,  it  is  clearly  evident 
that  the  given  power  is  not  a  perfect  square ;  and  since  the 
next  figure  of  the  root  is  5,  we  increase  the  last  figure  just 
found  by  1  and  write  —  after  the  root. 

In  this  problem  there  are  seven  periods — four  in  the  whole 
number  and  three  in  the  decimal ;  hence,  there  will  be  seven 
figures  in  the  root,  four  figures  constituting  the  whole 
number  and  three  figures  the  decimal  of  the  root.  Hence, 
4/9,000,099.4009  =  3,000.017  —  .  Ans. 

(5)  Beginning  at  the  decimal  point  and  pointing  off  into 
periods  of  three  figures  each,  we  have  .327'680,  annexing  a 
cipher  to  complete  the  right-hand  period.     Since  the  num- 
ber is  entirely  decimal,  the  root  is  entirely  decimal.    Neglect- 
ing the  decimal  point    for  the   present,   we   find   the  cube 
root  of  327,680.     Referring  to  the  table  in  Art.  3O,  327,680 
lies  between  314,432,  the  cube  of  68,  and  328,509,  the  cube  of 
69 ;  the  first  two  figures  of  the  root  are,  therefore,  68.     The 
first  difference  is  328,509  —  314,432  =  14,077;  the  second^dif- 
erence   is   327,680  —  314,432  —  13,248;   and  13,248-4-14*077 
=  .94+.      Therefore,  the  first  four  figures  of  the  root  are 
6894.     Locating  the  decimal  point  according  to  Art.  32,  we 
have  the  |/.  32768  =  .6894+.     Ans. 

1  4  0  7  7  )  1  3  2  4  8.0  0  (  .9  4+ 

126693 

328509  327680  5787 0 
314432  314432  56308 
~~1  4  0  7  7  13248  1562 

(6)  Instead  of  finding  the  cube  root  of  2,  we  annex  a 
cipher  period  and  find  the  cube  root  of  2,000,  in  order  to 
obtain  two  figures  of  the  root  from  the  table  (see  Art.  32). 
Referring  to  the  table,  2,000  lies  between  1,728,  the  cube  of 
12,  and  2,197,  the  cube  of  13;   the  first  two  figures  of  the 
root  are,  therefore,  12.     The  first  difference  is  2,197  —  1,728 
=  469;    the    second    difference    is  2,000—1,728  =  272.   and 
272  -^  469  =  .58  — .     Therefore,  the  first  four  figures  of  the 


§  2  ARITHMETIC.  17 

root  are  1258.  Since  there  is  only  one  period  in  the  num- 
ber, the  whole-number  part  of  the  root  will  contain  only 
one  figure;  hence,  the  root  is  1.258  —  .  We  need  one  more 
figure,  however;  hence,  proceeding  as  in  Art.  35,  2  -f-  1.258 
—  1.58983;  1.58983  -f-  1.258  =  1.26377;  1.258  -f  1.258 
+_1. 26377  =3.77977;  3.77977  -=-  3  =  1.25992.  Therefore, 
j/2  =  1.2599  to  four  decimal  places. 

2197  2.000  46  9)  272.000  (.5  79 -K  or  .5  8- 
1728  1728  2345 


469     272  3750 

3283 


4670 
4221 

1.2  5  8  )  2.0  0  0  0  0  0  0  0  0  (  1.5  8  9  8  2  5+,  or  1.5  8  9  8  3  — 

1258 


7420 
6290 

11300    1.258)1.58983000(1.26377 
10064  1258 


12360  3318 

11322  2516 


10380  8023 

10064          7548 


3160          4750 
2516  3774 


6440          9760 
6290'         8806 


9540 

1.2  5  8  8806 

1.2  5  8 
1.2  6  3  7  7 


3  )  3. 7  7  9  7  7 

1.2  5  9  9  2,  or  1.2  5  9  9  Ans. 


18  ARITHMETIC.                                   §  2 

(7)    (a)  (b) 

I  4/l'2  3.2  1  =  11.1  Ans.   1      /I'l  4.9  2'1  0  =10.72  + 

II  II  Ans- 

20  ~23  2001492 
_1        2J.  7      1449 

21  ~221  20T       ~4310 
_1           221  _  7            4284 
2~20  2T40                2~6 

1  2 


221  2142 

(8)  (a)  Since  the  number  is  entirely  decimal,  the  root  is 
entirely  decimal.  Adding  two  ciphers  to  complete  the  second 
period,  and  neglecting  the  decimal  point,  we  find  the  cube 
root  of  6,500.  Referring  to  the  table,  6,500  lies  between 
5,832,  the  cube  of  18,  and  6,859,  the  cube  of  19;  the  first  two 
figures  of  the  root  are  18.  The  first  difference  is  6,859 
—  5,832  =  1,027  ;  the  second  difference  is  6,500  —  5,832  =  668  ; 
and  668  -*-  1,027  =  .65+.  Therefore,  the  first  four  figures  of 
the  root  are  .1865+.  Since  five  figures  are  required,  we 
proceed  as  in  Art.  35;  .0065  -r-  .1865  =  .03485254;  .03485254 
-=-.1865=.  186876;  .1865  +.1865  +.186876  =.559876;  .559876 
-4-  3  =  .186625,  or  .18663  —  ,  correct  to  five  figures. 

6859     6500      1027)66  8.0  0(.  6  5+ 
5832     5832  6162 

1027       668  5180 

5135 

.1865).  006500000000  (.0348525  4+ 
5595 
9050 
7460 
1590  0 
14920 
9800 
9325 


4750 
3730 
10200 
932  5 
^T7l>0 
7  4  6Q 
1290 


ARITHMETIC.  19 

.1  8  6  5  ).0  3  4  8  5  2  5  4  0  0  (.1  8  6  8  7  6  + 
1865 


16202 
14920 

12825 

11190 

16354 
14920 


14340 
13055 

T¥iio 

•18G  5  1  1  1  QO 

.186876 

1660 


3  ).5  5  9  8  7  6 

.18662  5+,  or  .18663—.  Ans. 

(b)  Since  the  number  is  entirely  decimal,  the  root  is 
entirely  decimal.  Pointing  off  into  periods,  we  obtain .  000'021. 
The  figures  of  the  root  will  be  the  same  as  for  the  cube 
root  of  21.  Hence,  consider  the  given  number  as  21  and 
annex  a  cipher  period  so  that  two  figures  of  the  root  may  be 
obtained  from  the  table;  in  other  words,  we  extract  the 
cube  root  of  21,000.  Referring  to  the  table,  21,000  lies 
between  19,683,  the  cube  of  27,  and  21,952,  the  cube  of  28; 
the  first  two  figures  of  the  root  are,  therefore,  27.  The  first 
difference  is  21,952  —  19,683  —  2,269;  the  second  difference 
is  21,000  —  19,683  =  1,317 ;  and  1,317  ^  2,269  =.58+.  Hence, 
the  first  four  figures  of  the  root  are  2758.  Locating  the 
decimal  point  according  to  the  principle  given  in  Art.  32, 
the  root  is  .02758+.  The  result  does  not  agree  with  the 
printed  answer.  Hence,  proceeding  as  directed  in  the  note 
following  example  3  (see  Examination  Questions),  we  apply 
the  method  described  in  Arts.  35  and  36.  .000021  -f-  .02758 
=  .000761421;  .000761421  -r-  .02758  =  .027607;  .02758 
+  .02758  +  .027607  =  .082767;  .082767  -f-  3  =  .027589,  or 
.02759  —  ,  correct  to  four  significant  figures. 
H.  8.  I.— 29 


20  ARITHMETIC.  §  2 

21952     21000     2  2  6  9  )  1  3  1  7.0  0  (.5  8  + 
19683      19683  11345 


2269       1317  18250 

18152 


.  02758).  00002100000000(.  00076142  1-f- 
19306 


16940 
16548 

3920 

2758 


11620 
11032 


5880 
5516 

3640 

2758 


,0  2  7  5  8  )  .0  0  0  7  6  1  4  2  1  0  0  (  .0  2  7  6  0  7+ 
5516 


20982 
19306 


16761  .02758 
16548  .02758 
.027607 


21300     

19306    3). 082767 

1994       .0  2  7  5  8  9,  or 

.02759-  Ans 


2  ARITHMETIC.  21 

(9)      11.7  :  13  ::  20  :  x  The  product  of  the  means 

11.7  x  =  13  X  20  equals    the    product    of    the 

11.7  x  =  260  extremes. 

_   260 

~~  1  1.7  )  2  6  0.0  0  0  (  2  2.2  2+     Ans. 
234 

260 
234 

260 

234 


260 
234 

~2~6 


(10)     (a)  20  +  7  :  10  +  8  ::  3  :  x. 

27  :  18  ::  3  :  x 
27  x  =  18  X  3 
27  x  =  54 

54 
x  =  —  =  2.     Ans. 


(b)  122  :  1002  ::  4  :  x  12        100 

144  :  10,000  ::  4  :  x  12  100 


_  40,000 

~ 


144)  4  0  0  0  0.0  (  2  7  7.7+     Ans. 
288 


1120 
1008 

1120 
1008 


22  ARITHMETIC.  §  2 

(11)     (*)—==  Jj    is    equivalent    to    4:*::7:21.      The 

product  of  the  means  equals  the  product  of  the  extremes 
Hence, 

7  x  =  4  X  21 

7*=  84 


0    A 

x  =  — ,  or  12.     Ans 


In  like  manner, 


(b)  ^7  =  r«  is  equivalent  to  x  :  24  ::  8  :  16. 

16  x  =  24  X  8 
16;r=  192 
192 
=  T6~  = 

2          x 

(c)  —  =  —  is  equivalent  to  2  :  10  ::  x  =  100. 

10^=  2  X  100 
10  .ir  =  200 

200 
x  =  —  =20.     Ans. 

15       60 

(d)  —  =  —  is  equivalent  to  15  :  45  ::  60  :  x 


=  45  X  60 
15  x  —  2,  700 

2,700 
* 


A 

=  180.     Ans. 
15 


(e)     ——  =  ~  is  equivalent  to  10  :  150  ::  x  :  600. 
150       600 

150 x—  10  X  600 
150  x=  6,000 

6,000 
^=--=40.     Ans. 


§  2  ARITHMETIC.  23 


(12)     3£  ft.  =  3.5  ft.,  since  \  -  .5. 
6|  ft.  =  6.  75  ft.  ,  since  f  =  .  75. 


Consider  the  question  "What  do  we  wish  to  find?"  In 
this  case  it  is  "  pounds."  We  know  that  a  piece  of  shafting 
3.5  ft.  long  weighs  37.45  lb.,  and  we  wish  to  know  the  weight 
of  a  piece  of  shafting  6|  ft.  long.  It  is  evident  that  the 
weight  of  a  piece  of  shafting  6.75  ft.  long  bears  the  same 
relation  to  the  weight  of  a  piece  3.5  ft.  long  that  6.75  ft. 
bears  to  3.5  ft.  Letting  x  occupy  any  place  in  the  propor- 
tion, we  have  the  following,  the  value  of  x  being  the  same 
in  each.  Thus, 

(a)  3.5  ft.  :  6.75  ft.  ::  37.45  lb.  :  x  lb., 

6.75  X  37.45      252.7875 

or  x  =  —  —  =  —     -  —  =  72.225  lb.     Ans. 

o.o  o.5 

(b)  6.75  ft.  :  3.5  ft.  ::  x  lb.  :  37.45  lb., 

6.75X37.45      252.7875 

orx  =  ---  —  --  =  —  —  —  =  72.225  lb.     Ans. 
o.o  o.5 

(0     x  lb.  :  37.45  lb.  ::  6.75  ft.  :  3.5  ft., 

6.75  X  37.45      252.7875 
orx  =  —     —  -  --  =  —  -—  —  =  72.225  lb.     Ans. 

O.O  O.O 

(d)     37.45  lb.  :  x  lb.  ::  3.5  ft.  :  6.75  ft., 
6.75  X  37.45      252.7875 


3.5  3.5 


=  72.225  lb.     Ans. 


(13)  We  will  first  reduce  8  hr.  40  min.  to  minutes. 
8  hr.  -f-  40  min.  =  (8  X  60  min.)  +  40  min.  =  520  min.  In 
this  problem  we  are  required  to  find  "time."  We  know 
that  a  railway  train  runs  444  miles  in  520  minutes,  and  we 
want  to  know  how  long  it  will  take  it  to  run  1,060  miles 
at  the  same  rate  of  speed.  It  is  evident  that  the  time  it 
requires  to  run  1,060  miles  bears  the  same  relation  to  the 
time  it  takes  to  run  444  miles  that  1,060  miles  bears  to 
444  miles.  Letting  x  occupy  any  place  in  the  proportion, 
we  have  the  following,  the  value  of  x  being  the  same  in  each. 
Thus, 


24  ARITHMETIC.  §  2 

(a)     1,060  miles  :  444  miles  ::  x  min.  :  520  min., 
1,060  X  520       551,200 

-^44-       -444- 

1060  4  4  4  )  5  5  1  2  0  0.0  0  (  1  2  4  1.4  4+  min. 

520  ,          444 


21200  1072 

5300  888 


551200  1840 

1776 


640 
444 

1960 

1776 

1840 
1776 

6  4 

Reducing  1,241.44  min.   to  hours  by  dividing  by  60,  we 
have 

6  0  )  1  2  4  1.4  4  (  20  hr.  41.44  min.     Ans. 
120 


4  1 

(b]  444  miles  :  1,060  miles  ::  520  min.  :  x  min. 

x~  ~ — 444 — ™  =  1»341-44:  min-,  or  20  hr.  41.44  min.      Ans. 

(c]  x  min.  :  520  min.  ::  1,060  miles  :  444  miles. 

x  =  1>06?*  52°  —  1,241.44  min.,  or  20  hr.  41.4'4  min.     Ans. 

(d]  520  min.  :  x  min.  ::  444  miles  :  1,060  miles. 
1,060  X  520 


444 


=  1,241.44  min.,  or  20  hr.  41.44  min.     Ans. 


(14)     A    pump   discharging    135    gal.    per    min.    fills  the 
tank  in  38  min.     Therefore,  a  pump  discharging  1  gal.  per 


§  2  ARITHMETIC.  25 

min.  fills  it  in  135  X  38  min.      Hence,  a  pump   discharging 

-|   O  K     \s     O  Q 

85  gal.  per  min.  fills  it  in  —          —  =  60T6T  min.     Ans. 
oo 

1  35 
X    38 


1080 
405 

5  )  5130  (  6  0 
510 


—  -TV 

8  5~TT 

(15)  If  a  wheel  measuring  12.56  ft.  around  it  turns  520 
times,  a  wheel  measuring  1  ft.  around  it  turns  520  X  12.56 
times.  Hence,  a  wheel  measuring  15.7  ft.  around  it  turns 

520  X  12.56 

-^-^ =  416  times.     Ans. 

lo.  I 

520 

1  2.5  6 


3120 
2600 
1040 
520 

1  5.7  )  6  5  3  1.2  0  (416  times. 

628 


251 

157 

942 

942 


(16)     If  a  cistern  28  ft.  by  12  ft.  by  10  ft.  holds  798  bbl. 
of  water, 

798 
a  cistern  1  ft.   by  12  ft.  by  10  ft.  holds-—-  bbl.  ; 

/io 

798 

a  cistern  1  ft.   by  1  ft.  by  10  ft.  holds  —       —  bbl.  ; 

AO  X  1* 

798 

a  cistern  1  ft.   by  1  ft.   by  1  ft.  holds—  —  —-bbl. 

(CO  X  1<«  X  10 


26  ARITHMETIC.  §  2 

Therefore,   by  a  similar  course  of   reasoning,  a   cistern 
20  ft.  by  17  ft.  by  6  ft.  holds 

57 

w    ? 

•  798  x  20  x  17  X  6  _  ftp  x  ?0  X  17  X  0  _  969  _ 

28  x  12  x  10          $$  x  ;2  x  ;p     r  2~ = 

^        ?  Ans. 


5  7 
17 

399 

5  7 

2)969 
484£ 


MENSURATION  AND  USE  OF 
LETTERS  IN  FORMULAS. 


(1)     Substituting  for  D,  x,  B,  and  i  their  values, 
D  -  x      120  -  12       108 


A  line  between  two  numbers  signifies  that  the  one  above 
the  line  is  to  be  divided  by  the  one  below  the  line. 

(2)  Substituting  for  A,  //,  D,  and  x  their  values, 

A/i  +  D  _  (5  X  200)  +  120       1,000  +  120  _  1,120  _ 
2  Jt  :  +  6  (2  X  12)  +  6  24+6  30 

37^  +  D  =  37£  +  120  =  157i.     Ans. 

When  there  is  no  sign  between  the  letters,  multiplication 
is  understood. 

(3)  Substituting  for  A,  D,  /,  and  B  their  values, 


_      /      A  D  I         5  X  120  /600 

3  V  i&+  1-5  ~  V  (3.5  X  10) +  1.5  ~  V  36.5 

=  4/16.4383  =  4. 05+.     Ans. 

The  square  root  sign  extends  over  both  numerator  and 
denominator,  thus  indicating  that  the  square  root  of  the 
entire  fraction  is  to  be  extracted. 

(4)     Substituting  for  A,  B,  D,  and  h  their  values. 


_  (B  —  AY  —  \/h  +  2 B  +  A  _  (10  —  5)2  —  j/200  +  2x10  +  5 
A3  —  (!  +  £>)  53  —  (1  +  120) 

5'  -  i/225"       25  -  15 


125  —  121  4 

§3 


Ans- 


For  notice  of  copyright,  see  page  immediately  following  the  title  page. 


2  MENSURATION  AND  §  3 

(5)  When  one  straight  line  meets  another  straight  line, 
two  angles  are  formed  which  together  equal  180°.     Hence,  if 
one  of-  the  angles  =  152°  3',  the  other  angle  =  180°  —  152° 
3',  or 

180°  =  179°  60' 
subtracting,       152°     3' 

27°  57'     Ans. 

(6)  See  Arts.  26-28. 

(7)  See  Art.  41.     A  rectangle  with  the  same  area  would 
have  the  same  base  and  altitude. 

(8)  Since  the  area  is  to  be  found  in  square  inches,'  the 
2£  feet  must  be  reduced  to  inches.     2£  ft.  =30  in.     Area 
=  30  X  1H  =  345  sq.  in.     Ans. 

(9)  It  will  take   1£  boards  to  reach   lengthways  of  the 
room.      Since  the  room  is  15  feet  wide  and  each  board  is 

5  incites  wide,  it  will  take  15  -f-  —  =  36  boards,  laid  side  by 

side,  to  extend  across  the  width  of  the  room.     Hence,  num- 
ber of  boards  required  =  36  X  1£  =  54.     Ans. 

(10)  The  total  area  of  the  floor  of  the  station  =  55  X  58  ft. 
=  3,190  sq.   ft.  —  25  X  26  ft.  =  650  sq.   ft.,  the   area  repre- 
sented by  the  lower  right-hand  corner  of  the  figure.     Hence, 
total  area  of  floor  =  3,190  —  650  =  2,540  sq.  ft. 

From  this  we  have  to  deduct  the  following  areas: 
2  boilers  =  2  X  8  X  19  =  304  sq.  ft. 
Feed-pump  =  2|  X  5  =  12.5  sq.  ft. 
2  engines  =  2  X  4|  X  10  =  90  sq.  ft. 
2  dynamos  =  2  X  5£  X  6£  =  71.5  sq.  ft. 

Switchboard^      1Q  *  3'5     =      2. 92  sq.ft. 


480.92sq.  ft. 

The  unoccupied  floor  space,  therefore,  equals 
2,540  -  480.92  =  2,059.08  sq.  ft.     Ans. 

(11)     A  triangle  with  three  equal  angles  has  three  equal 
sides,  and  is  therefore  an  equilateral  triangle. 


§  3  USE  OF  LETTERS  IN  FORMULAS.  3 

(12)  A   triangle  with  two    equal   angles  has  two  equal 
sides,  and  is  therefore  an  isosceles  triangle. 

(13)  The  sum  of  the  three  angles  in  any  triangle  =•  2  right 
angles,    or   180°.     In  the   given   triangle,  the  sum    of   two 
angles  =  23°  +  32°  32'  =  55°  32',  and  the  third  angle  =  180° 
—  55°  32',  or 

180°  =  179°  60' 
subtracting,         55°  32' 

124°  28'     Ans. 

(14)  In  Fig.  I  we  have  the  propor- 
tion A  D  :  D  E  ::  A  B  :  B  C,  in  which 
AD=10  in.,  A  £  =  '24:  in.,    and  BC 
=  13$  in.,  to  find  D  E. 

Substituting  the  given  values, 
10  :  DE  ::  24  :  13$,  or 

n  -      10  X  13.5  . 

DE  =  —  —  =  5.625  in.     Ans. 

FIG.  I. 

(15)  A  line  drawn  diagonally  from  one  corner  to  the  op- 
posite one  would  form  the  hypotenuse  of  a  right  triangle, 
whose  two  sides  are  39  and  52_feet.  .  By  rule  6,  Art.  58, 

the  length  of  the  diagonal  =  /52s  +  39'  =  65  ft.     Ans. 

(16)  See  example,  Art.  64.     The  process  is  simply  to 
find  one  of  the  angles  of  the  polygon,  and  then  to  divide  it 
by   2.      By  rule   1O,   Art.    64,   one  of    the   interior  angles 
_  180  X  (8  -  2)  _  1350^ 

8 

(17)  Since  this  is  a  regular  hexagon,  it  may  be  inscribed 
in  a  circle  (Fig.  II),  and  the  radius  of  the  inscribing  circle 

will  be  equal  to  one  side  of  the  hexagon. 
Since  the  diameter  E  F  •=  2  inches,  the 
radii  A  B  and  A  C,  and  the  side  B  C 
each  =  1  inch,  and  the  triangle  ABC 
is  equilateral.  Draw  the  line  A  D  per- 
pendicular to  the  side  B  C\  it  will 
bisect  B  C.  Then,  in  the  right-angled 
triangle  ADB,AB=Y  and  B  D  =  $', 


MENSURATION  AND 


to  find  A  D.  According  to  rule  7,  Art.  59,  A  D  =  4/1"  —  .5" 
=  I/.75  =  .866".  Hence,  the  distance  between  two  opposite 
sides  of  the  hexagon  —  A  D  X  2  =  .866  X  2  =  1.732".  Ans. 

(18)     In  Fig.   Ill,  we  have  the  pro- 
portion B  I :  H  I\\H  I\  I  A,  in  which 

B  1=  6  and  HI-  |of  HK=  —  =  9. 
tf  2 

Substituting,    6  :  9  ::  9  :  I  A,    or   I A 

81 

=  —-  =  13. 5  in.   Hence,  the  diameter  A  B 
o 

FIG.  m.  =  I A -\-BI-  13.5  +  6  =  19.5  in. 

Ans. 

(19)  One  mile  =  5,280  feet.     The  circumference  of  the 

wheel  in  feet  =  SLSJLlii!  =  18.8496.      (See  rule  12,  Art. 
1/c 

77.)  Number  of  revolutions  in  going  1  mile  =  5,280 
-f-  18.8496  =  280.112.  Ans. 

(20)  Using  rule   15,  Art.   8O,  area  =  diameter  squared 
X  .7854.     6.06"  =  36.7236;  36.7236  X  .7854  =  28.8427  sq.  in. 

Ans. 

(21)  Since  the  radius  of  the  circle  =  6  in.,   its  diameter 
=  12  in.,  and  its  circumference  =  12  X  3.1416  =  37.6992  in. 
There  are  360°  in  the  circumference,  and  the  length  of  an 

arc  of  12°  =  37.6992  X  ~  =  1.25664  in.     Ans. 
ooO 

(22)  The  area  of  a  circle  15  in.  in  diameter  =  15"  X  .7854 
=  176.715    sq.    in.      Hence,    the    area    of   a  sector    of   this 

101       2  208  937 
circle    whose    angle    is     12^°  =  176.715  X  TJ~|  =  — 

ouO  ouO 

=  6.1359  sq.  in.     Ans.      (See  rule  17,  Art.  82.) 

(23)  (a)  The  side  of  a  square  whose  area  =  103.8691  sq.  in. 
=  1/103.8691  =  10.1916  in.     Ans. 

(b)     By  rule  16,  Art.  81,  the  diameter  of  a  circle  having 

/103.8691 

the  same  area  =A/  — f)QRA          H$  m-     Ans. 


§  3  USE  OF  LETTERS  IN  FORMULAS.  5 

(f)  Perimeter  of  the  square  =  10.1916  X  4  =  40.7664  in.  ; 
circumference  of  the  circle  =  11.5  X  3.  1416  =  36.1284  in.; 
difference  =  40.  7664  —  36.  1284  =  4.  638  in.  Ans. 

(24)  The  perimeter  of  the  base  =  4  X  6  —  24  in.  =  2  ft. 
Convex  area  —  2  X  12  =  24  sq.  ft.  The  area  of  the  bases 
is  found  as  follows:  In  Fig.  IV,  A  B 
=  4  in.  and  A  C  =  2  in.  ;  since  this  is  a 
regular  hexagon,  A  O  =  A  B  =  4  in.  By 
rule  7,  Art.  59,  O  C  =  4/4*  -  2'  =  /12 
—  3.4641  in.;  area  of  triangle  AOB 


base  =  6.9282  X  6  =  41.5692;  and  the  area  FIG.  iv. 

of  both  bases  =  41.5692  X  2  —  83.1384  sq.  in.     This  reduced 

83  1384 
to  square  feet  =  ^  —  .5774.      Hence,   the  area  of   the 

entire  surface  of  the  column  is  24  +  .5774  =  24.5774  sq.  ft. 

Ans. 

(25)  The  cubical  contents  in  cubic  inches  =  area  of  base 
in    square    inches  X  altitude  in  inches.      The   area   of   the 
base  in  the  last  example  was  found  to  be  41.5692  sq.   in.  ; 
altitude  =  12  X  12  =  144  in.      Hence,    the  cubical  contents 
—  41.5692  X  144  =  5,985.9648  cu.  in.     Ans. 

(26)  This  example  is  solved  by  combinirig  the  rules  for 
the  circular  ring  (see  example,  Art.  81)  and  for  the  cylinder. 
To  obtain  the  area  of  one  end  of  the  tube,  we  have  4"  X  .7854 
=  12.5664  =  area    of    a   circle  4  inches  in    diameter;    3.73* 
X  .7854  =  10.9272  =  area  of  a  circle  3.73  inches  in  diameter; 
difference  =  12.5664  —  10.9272  —  1.6392  —  area  of  one  end  of 
the  tube.   The  cubical  contents  =  1.  6392  X  12  =  19.  6704cu.  in.  ; 
the  weight  =  19.6704  X  .28  =  5.5,  or  5|  Ib.     Ans. 

(27)  This  example  is  done  exactly  like  the  one  in  Art.  92, 
and  the  solution  is  given  here  without  explanation. 


(a)     In  the  formula  of  rule  18,  Art.  83,  np  r  -     —  -608, 
Ji  in  this  case  =  is,  and  D  =  60. 


MENSURATION  AND  §  3 

Substituting,  area  = 


4X324 


-  .  Q        4/3.333-.  608  =  432  X  4/2.725 

0  lo  o 

=  432  X  1.65  =  712.8  sq.  in.     This  reduced  to  square  feet 

=  712.8  -f-  144  =  4.95.  Hence,  the  steam  space  =  4.95 
X  16  =  79.2  cu.  ft.  Ans. 

(b)  Total  area  of  one  end  of  boiler  in  square  inches  =  GO2 
X  .7854  —  2,827.44.  From  this  is  to  be  subtracted  the  area 
of  the  tube  ends  and  of  the  segment  found  above. 

Area  of  ends  of  tubes  —  3.5a  X  .7854  X  64  =  G15.75  sq.  in. 

Area  of  segment  =  712.8    sq.  in. 

1,328.55  sq.  in. 

Area  of  water  space  =  2,  827.  44  —  1,  328.  55  =  1,  498.  89  sq.  in  . 

Contents  of  water  space  =  1,498.89  X  16  X  12  =  287,- 
786.88  cu.  in.,  and  287,786.88  -f-  231  =  1,245.83,  number  of 
gallons,  or  say  1,246  gal.  Ans. 

(28)  The  area  of  the  convex  surface  =  circumference  of 

base  X  \  slant  height  =  18.8496  X  -r-  =  94.248  sq.  in.      (See 

"A 

rule  SI,  Art.  97.)  The  area  of  the  entire  surface  —  94.248  sq. 
in.  -f-  the  area  of  the  base.  The  diameter  of  the  base 

1  R  R4-Qfi 

_  ICLM  tfo  _  6  hence    the  area  of  the  base  =  6'  X  .7854 

3.1416 

=  28.2744  (rules  13  and  15,  Arts.  78  and  8O)  ;  there- 
fore, the  area  of  the  entire  surface  =  94.248  +  28.2744 
=  122.5224  sq.  in.  Ans. 

(29)  Using  rule  22,  Art.  98,  volume  =  area  of  base  X 

-i  altitude  =  28.2744  X  |  —  84.8232  cu.  in.     Ans. 
o 

(30)  The  vat  has  the  form  of  an  inverted  frustum  of  a 
pyramid.     Area   of   larger   base  =  152  =  225    sq.    ft.  ;    area 
of   smaller  base  =  122  =  144  sq.   ft.      Hence,   by  rule   24, 
Art.     1O2,    the    contents    of     the    vat    in    cubic    feet  — 


(225  +  144  +  4/225  X  144)       =  (369  +  180)  X  ~  =  549  X 

o  o  o 

=  2,013  cu.  ft.     This  should  be  reduced  to  cubic  inches  by 


§  3  USE  OF  LETTERS  IN  FORMULAS.  7 

multiplying  by  1,728,  the  number  of  cubic  inches  in  a  cubic 
foot.  2,013  X  1,728  =  3,478,464  cu.  in.  Since  there  are 
231  cubic  inches  in  a  gallon,  the  number  of  gallons  that  the 

q  470   AC <A 

vat  will  hold  =    '    *°1        =15, 058. 29.     Ans. 

/vol 

(31)  The  pail  is  in  the  form  of  a  frustum  of  a  cone. 
Area  of  larger  base  =  12"  X  .7854  =  113.0976  cu.  in.  Area 
of  smaller  base  =  63. 6174  cu.  in.  Hence,  the  contents  in 
cubic  inches  = 


11 


(113.0976  +  63.6174  +  /113.0976  X  63.6174)  X  -_• 


=  (176.715  +  4/7,  194.  9753)  ~  =  (176.715  +  84.8232)  X 

o  o 

=  261.5382  X  ^  =  958.9734. 
o 

The  contents  of  the  vat  in  cubic  inches  were  found  in 
the  last  example  to  be  3,478,464.  Hence,  the  number  of 
pails  of  water  required  to  fill  the  vat  =  3,478,464-^-  958.9734 
=  3,627.28.  Ans. 

(32)  (a)  By   rule    25,    Art.    1O4,   area   of    the    surface 
=  22.5*  X  3.1416  =  506.25  X  3.1416  =  1,590.435  sq.  in.     Ans. 

(/;)  Using  rule  26,  Art.  1O5,  the  cubical  contents 
=  the  cube  of  the  diameter  X  .5236  =  11,390.625  X  .5236 
=  5,964.1313  cu.  in.  Ans. 

(33)  (a)    Given    O£=^,  or  8    inches,    and    0'A=-£t 

or  G£   inches,    to    find    the    volume,   area,    and  weight  (see 
Fig.  V): 

Radius    of    center    circle    equals      "t"    '   ,    or    71    inches. 

a 

Length  of  center  line  =  2  X  3.1416  X  7± 
—  45.5532  inches. 

The  radius  of  the  inner  circle  is 
6£  inches  and  of  the  outer  circle  8  inches  ; 
therefore,  the  diameter  of  the  cross- 
section  on  the  line  A  B  is  1|  inches. 
Then,  according  to  rule  27,  Art.  1O6, 


8  MENSURATION.  §  3 

the  area  of  the  ring  is  1|  X  3.1410  X  45.553  =  214.065  square 
inches.  Ans. 

Diameter  of  cross-section  of  ring  —  1£  inches. 

Area  of  cross-section  of  ring  =  (H)2  x  .7854=  1.76715  sq. 
in.  Ans. 

By  rule  28,  Art.  1O7,  volume  of  ring  =  1.70715  X  45.553 
=  80.499  cu.  in.  Ans. 

(b)     Weight  of  ring  =  80.499  X  .261  =  21  Ib.     Ans. 


PRINCIPLES  OF  MECHANICS. 


(1)  Broadly    speaking,    mechanics    is    the    science   that 
treats  of  forces  and  their  effects  on  material  bodies.     See 
Art.  1. 

(2)  See  Arts.  2,  3,  and  4. 

(3)  Solid,  liquid,  and  gaseous.     See  Arts.  6,  7,  and  8. 

(4)  General  properties  are  such  as  are  common  to  all 
substances;  special  properties  are  such  as  are  possessed  by 
certain  substances  only.     See  Arts.  12  and  24. 

(5)  (a)  Motion  is  that  condition  of  a  body  that  causes  it 
to  change  its  position  in  relation  to  some  other  body.     See 
Art.  3O. 

(fr)     Velocity  is  the  rate  of  motion,  that  is,  the  distance 
passed  through  in  a  unit  of  time.     See  Art.  32. 

(c)  Uniform  velocity  means  passing  over  equal  distances 
in  equal  intervals  of  time.     See  Art.  32. 

(d]  Variable  velocity  means  passing  over  equal  distances 
in  unequal  intervals  of  time,  or  passing  over  unequal  dis- 
tances in  equal  intervals  of  time.     See  Art.  33. 

(6)  (a]  Acceleration  is  the  rate  of  change  of  velocity. 
See  Arts.  35  and  57. 

(b)  Retardation  denotes  the  rate  of  decrease  of  velocity. 
See  Arts.  36  and  58. 

(c)  The  average  velocity  is  that  uniform  velocity  that 
carries  the  body  over  the  same  distance  in  the  same  time  as 
the  variable  velocity.     See  Art.  37. 

§4 
H.    8.    I.— SO 


2  PRINCIPLES  OF  MECHANICS.  §  4 

(7)  Since  1  day  contains  24  hours,  the  total  number  of 
hours  consumed  in  the  trip  was    6  X  24  +  16  =  160   hours. 
The    average   speed    per   hour   was,   therefore,    3,240-^-160 
=  20£  mi.     Ans.     See  Art.  38. 

(8)  Reducing  the  25,000  miles  to  feet,    we  have  25,000 
X  5,280  =  132,000,000    feet.     The  time  in  minutes  would, 
therefore,  be  132,000,000  -h  3,000  =  44,000  minutes.      Redu- 
cing the  time  to  days,   hours,   and  minutes,  we  get  30  da. 
13  hr.  20  min.     Ans.     See  Art.  4O. 

(9)  By  the  effects  it  produces  on  matter.     See  Art.  41. 

(10)  The  point  of  application,   the   direction  of  action, 
and  the  magnitude  of  each  force   must   be   known.      See 
Art.  44. 

(11)  See  Art.  46. 

(12)  Inertia  is  that  property  of  a  body  by  virtue  of  which 
the  body  always  tends  to  remain  in  the  particular  state  of 
rest  or  motion  that  it  has  at  the  moment  considered.     See 
Art.  48. 

(13)  See  Arts.  52  and  53. 

(14)  The  weight  of  a  body  is  proportional  to  the  force  of 
gravity  ;  and  since  the  force  of  gravity  is  different  for  differ- 
ent localities,  the  weight  of  any  body  will   differ  likewise. 
See  Art.  56. 

(15)  By  rule  4,  Art.  6O, 


(16)  See  Art.  62. 

(17)  The  constant  opposing  force  is  30  X  15  =  450  pounds. 
30  tons  =  30  X  2,000  =  60,000  pounds.     The  force  /  to  pro- 
duce the  required  acceleration  will,  by  rule  5,  Art.  63,  be 

-  —  a  —      '         X  3  =  5,597+  pounds.     Hence,  the  total  force 
g  o2.1o 

required  will  be  5,597  +  450  ==  6,047  Ib.     Ans. 


PRINCIPLES  OF  MECHANICS.  ! 

5,280 


(18)     1    mile  per  hour  =  5, 280  feet   per   hour. 


60  X  60 
=  f  I  =  1-j-V  feet   per   second.       Momentum,    according   to 

Art.  67,  =  ^r£?  X  ??  =  2,  736.  3  Ib.     Ans. 
o/o.  lo         15 

(19)      (a)  Work  is  the  overcoming  of  a  resistance  through 
a  Certain  distance.     See  Arts.  69,  7O,  and  71. 

(b)  Power  is  the  rate  of  doing  work.      See   Arts.    72 
and  73. 

(c]  Energy  denotes  ability  to  do  work.     See  Arts.   74 
and  75. 


(21)  The  kinetic  energy  of  the  body  by  rule  7,  Art.  76, 

is  ^L  -  6'432  *  Q0'  =  360,  000  foot-pounds.     This  amount  of 
&  g         &  X  dJ8«  ID 

work  must  be  done  in  3  minutes.     Hence,  the  horsepower 

required  to  stop  the  body  will  be  360'000  _=_  33,000  =  3T7T  H.  P. 

o 

Ans. 

(22)  Zero.     See  Art.  86. 

(23)  Friction  is  the  resistance  that  a  moving  body  encoun- 
ters from  the  surface  of  another  body  along  which  or  through 
which  it  moves.     See  Art.  88. 

(24)  According  to  Art.  89,  the  coefficient  of  friction  will 

-06.     Ans. 


(25)  (a)  From  Table  I  we  find  the  coefficient  of  friction 
to  be  .16.      Hence,    the   total   friction  will   be   1,000  X  .16 
=  160  Ib.     Ans.     See  Art.  92. 

(b)     {-£§•==  1.6  Ib.  per  sq.  in.     Ans. 

(26)  The  center  of  gravity  of  a  body  is  a  point  at  which 
the  whole  weight  of  the  body  may  be  considered  as  concen- 
trated.   This  point  may  be  either  inside  or  outside  the  body. 
See  Art.  98. 

(27)  See  Art.  99. 


4  PRINCIPLES  OF  MECHANICS.  §  4 

(28)  Suspend  the  body  successively  from  two  different 
points  and  find  the  point  of  intersection  of  two  plumb-lines 
from  the  points  of  suspension.     The  center  of  gravity  will 
be  on  the  line  passing  through  the  point  of  intersection  and 
perpendicular  to  the  plumb-lines.     See  Art.  1O3. 

(29)  (a)    Centrifugal   force  is  that  force  that  tends   to 
pull  a  revolving  body  away  from  the  point  about  which  the 
body  revolves.     See  Art.  1O4. 

(If)  Centripetal  force  is  that  force  that  tends  to  pull  a 
revolving  body  toward  the  point  about  which  it  revolves. 
See  Art.  1O5. 

(30)  The  centrifugal  and  centripetal  forces  of   a  given 
revolving  body  are  always  equal  and  opposite.    See  Art.  1O4. 

(31)  Applying  rule  14,  given  in  Art.  1O6, 

F=  .00034  X  10  X  6  X  60'  =  73.44  Ib.     Ans. 

(32)  See  Arts.  11O,  111,  and  112. 

(33)  The  forces  acting  on  a  body  at  rest  are  in  equilib- 
rium.    See  Art.  1O9. 

(34)  As  long  as  the  line  of  direction  falls  within  the  base, 
the  body  will  stand ;  if  it  falls  outside  the  base,  the  body  will 
fall.    In  the  former  case  the  forces  will  be  in  equilibrium ;  in 
the  latter  they  will  not  be  in  equilibrium.     See  Art.  113. 


MACHINE    ELEMENTS. 


(1)  (a)  A  lever  is  a  rigid  bar  or  rod  capable  of  being 
turned  about  a  pivot. 

(b]  The  weight  arm  is  that  part  of  the  lever  between  the 
fulcrum  and  the  weight. 

(c]  The  force  arm  is  that  part  of  the  lever  between  the 
fulcrum  and  the  force. 

(d~)     The  fulcrum  is  the  point  or  pivot  about  which  the 
lever  turns.     See  Arts.  1  and  2. 

(2)  The  product  of  the  weight  and  the  weight  arm  must 
be  equal  to  the  product  of  the  force  and  the  force  arm.     See 
Art.  2. 

(3)  See  rule  1,  Art.  3. 

1 2  V  20 

(4)  —  =  2.4  in.     Ans.     See  Art.  3. 
JLUO 

(5)  Solid  pulleys  and  split  pulleys.     See  Art.  9. 

(6)  Split  pulleys  can  be  more  easily  put  on  and  removed 
from  shafts  than  solid  pulleys.     See  Art.  9. 

(7)  As  explained  in  Art.  11,  the  belt  climbs  toward  the 
highest  part  of  the  pulley;  and  since  in  a  crowned  pulley 
the  highest  portion  is  in  the  center,  the  belt  will  tend  to  stay 
on  the  pulley. 

(8)  "  Balancing  pulleys  "  is  the  operation  of  making  oppo- 
site sides  of  pulleys  equal  in  weight,  so  that  the  centrifugal 

§5 


2  MACHINE  ELEMENTS.  §  5 

forces  of  the  opposite  sides  of  a  revolving  pulley  will  be 
equal.  A  pulley  is  balanced  by  first  finding  which  side  is 
the  lighter  and  then  adding  weights  to  the  lighter  side  until 
it  is  as  heavy  as  the  other.  See  Art.  12. 

(9)  The  "driver"  is  the  pulley  that  imparts  motion  to 
the  belt.     The  "driven"  is  the  pulley  that  receives  motion 
from  the  belt.     See  Art.  14. 

(10)  Applying  rule  3,  Art.  15, 

^      6  X  1,800 
77  =  —600—  =  18m-     AnS" 

(11)  Applying  rule  6,  Art.  18, 

N  =  5X  I'600  =  666|  rev.  per  min.     Ans. 
l/o 

(12)  Applying  rule  7,  Art.  23, 

B  =  3i  (^-          +  2  X  40  =  93  ft.     Ans. 


(13)  From  Table  I  we  find  that  the  allowable  effective 
pull  Cfor  an  arc  of  contact  of  120°  is  28.8  pounds.'    Apply- 
ing rule  8,  Art.  27, 

,,,       33,000  X  40 

W=  1,650  X  28.8  =  OT.7.«y»8.n.     Ans. 

(14)  Applying  rule  9,  Art.  28, 
^=28-8X33a|800X01-2°°=29.3+H.P.     An, 

(15)  Applying  rule  1O,  Art.  31, 


(16)  The  hair  or  grain  side  is  commonly  considered  to  be 
the  proper  side  to  be  in  contact  with  the  pulley  face.     For 
reasons  see  Art.  32. 

(17)  No;  rosin  makes  the  belt  gummy  and  causes  it  to 
crack.     See  Art.  32. 


§  5  MACHINE  ELEMENTS.  3 

(18)  Pulleys  run  out  of  true;  they  should  be  turned  true. 
Pulleys  out  of  line ;  they  should  be  lined  up.     Belts  some- 
times flap  if  they  are  run  at  speeds  over  4,000  feet  per  min- 
ute; perforating  the  belt  with  a  series  of  small  holes  is  said 
to  remedy  the  defect  in  this  case.     Lack  of  steadiness  in 
running;  take  steps  to  insure  steady  running.     A  defective 
joint;  unlace  the  joint  and  relace  it  properly.     Too  great  a 
distance  between  pulleys ;  reduce  the  distance  if  possible  or 
substitute  a  wider  belt.      See  Art.  33. 

(19)  Lacing,    sewing,     riveting,     and    cementing.     The 
cemented  joint  is  considered  the  best. 

(20)  See  Art.  38. 

(21)  A  wheel  that  imparts  motion  to  another  is  called  a 
"driver."     A  "follower"  is  a  wheel  that  receives  motion 
from  another. 

(22)  Letting     W  represent    the    weight    we    have,     by 
rule  13,  Art.  39, 

50  X  90  X  30  X  36  =  IV  X  30  X  12  X  20, 
or  4,860,000  =  7,200  W. 

,,,      4,860,000 
Hence,  W  =    'y  ^Q      =  675  Ib.     Ans. 

(23)  (a)  Circular  pitch  is  the  distance  measured  along 
the  pitch  circle  from  a  point  on  one  tooth  to  the  correspond- 
ing point  on  the  next  tooth.     See  Art.  47. 

(b)     Diametral  pitch  is  the  number  of  teeth  per  inch  of 
pitch  diameter.     See  Art.  47. 

(24)  The  epicycloidal  and  the  involute.     See  Arts.   51 
and  52. 

(25)  Involute  teeth  are  stronger  and  their  action  is  more 
satisfactory.     See  Art.  52. 

(26)  Applying  rule  13,  Art.  54, 

n      1.152  X  60 

D=      3.1416      =22m- 


MACHINE  ELEMENTS.  §  5 


(27)  Applying  rule  15,  Art.  56, 

D      3.1410  X  30  '• 

P=—  —  =  1.5 

60 

(28)  Applying  rule  17,  Art.  59, 


D      3.1410  X  30  '•*. 

P=—  —  =  1.57m.     Ans. 

60 


Ans. 


(29)  Applying  rule  21,  Art.  63, 

7V=  7.5  X  8  —  2  =  58  teeth.     Ans. 

(30)  A  fixed  pulley  has  a  stationary  block.     A  movable 
pulley  has  a  movable  block.     See  Arts.  71  and  72. 

(31)  Calling  W^the  weight  that  can  be  raised,  we  have, 
by  applying  rule  27,  Art.  74, 

W=  150  X  14  =  2,100  Ib.     Ans. 

(32)  The  force  that  must  be  applied  is  greater  than  if 
there  were  no  frictional  losses.     See  Art.  75. 

(33)  Applying  rule  28,  Art.  75, 

100  X  1,500 

/„=  -      —  —312.5  Ib.     Ans. 
Ja        2  X  4  X  60 

(34)  With  the  Western  differential  pulley  block  a  much 
greater  weight  can  be  raised  with  a  given  force  than  with 
the  ordinary  pulley  ;  and  the  load  can  be  stopped  anywhere 
by  ceasing  to  pull  on  the  chain.     See  Art.  78. 

(35)  Applying  rule  3O,  Art.  82, 

p=  60Q  *  12  =  102.  8+,  say  103  Ib.     Ans. 

(36)  Applying  rule  31,  Art.  83, 


lo 


(37)     First  applying  rule  34,  Art.  89,  we  find  the  factor 
by  which  the  theoretical  weight  is  to  be  multiplied. 

Thus,  £==_  =  A. 


§  5  MACHINE  ELEMENTS.  5 

Then  applying  rule  35,  Art.  9O,  to  obtain  the  theoretical 
weight,  we  find 

„,      6.2832  X  50  X  20 

W=—        — j —        —  =  37,699.2  pounds. 

Finally  applying  the  principle  of  Art.  92,  we  find  the 
probable  actual  weight  to  be 

37, 699. 2  X  TV  =  2, 356. 2  Ib.     Ans. 

(38)  (a)  The  velocity  ratio  is  the  ratio  between  the  dis- 
tance through  which  the  force  acts  and  that  through  which 
the  weight  moves.      See  Art.  1O3. 

($)     The  efficiency  is  the  ratio  of  the  actual  work  to  the 
theoretical  work.     See  Art.  1O5. 

9  3*ifi  9 

(39)  By    Art.    1O5,   the    efficiency    would   be  ^^ 

o7,o99./« 

=  .0625,  or  6£  per  cent.     Ans. 


MECHANICS   OF    FLUIDS. 


(1)  Hydrostatics  is  that  branch  of  mechanics  that  treats 
of  liquids  at  rest.     See  Art.  1. 

(2)  See  Art.  4. 

(3)  See  Art.  5. 

(4)  The  pressure  due  to  the  weight  of  water  acts,  like  any 
other  pressure  to  which  the  water  may  be  subjected,  in  all 
directions.     See  Art.  5. 

(5)  First  find  the   area  of  the   base.     This  equals  .7854 
X  33  =  7.0686  square  inches.      Multiplying  this  by  the  height 
of  the  water,  we  get  7.0686  X  8  =  56.5488  cubic  inches  as 
the  volume  of  a  column  of  water  whose  weight  is  equal  to 
the  pressure  on  the  base.     Therefore,  the  pressure  on  the 
base  =  56.5488  X  .03617  =  2.045+  Ib.     Ans.      See  Arts.  7, 
8,  and  9. 

(6)  See  Art.  12. 

(7)  The  water  stands   at   the  same  level  in   each  tube, 
because  it  is  then  in  equilibrium.      See  Art.  16. 

(8)  Because  of  the  resistance  of  the  air  and  the  friction 
in  the  pipe.      See  Art.  17. 

(9)  Specific  gravity  is  the  ratio  between  weights  of  equal 
volumes  of   any  substance   and  water.     Thus,   the  specific 
gravity  of  iron  is  the  ratio  between  the  weight   of  a  given 
volume  of  iron  and  the  weight  of  an  equal  volume  of  water. 
See  Art.  23. 

§6 


2  MECHANICS  OF  FLUIDS.  §  6 

(10)  Using   the    principle   of   Art.    25    and   taking    the 
specific  gravity  from  Table  I,  we  have  for  the  weight  of  the 
aluminum  62.42  X  2.50  X  100  =  15,605  Ib.     Ans. 

(11)  A  body  immersed  in  a  liquid  will  be  buoyed  up  with 
a  force   equal  to   the  weight  of  the  liquid  displaced.     See 
Art.  26. 

(12)  See  Art.  29. 

(13)  (a)   Hydrometers  are  instruments  for  determining 
the  specific  gravity  of  liquids  and  of  some  forms  of  solids. 

(b]  Hydrometers  of  constant  weight  and  hydrometers  of 
constant  volume.  See  Art.  3O. 

(14)  Applying  rule  5,  we  have 

.7854  X  I2  X  400 

n  =  —  —  =  16.32  gallons. 

iy.  /co 

As  there  are  60  minutes  in  1  hour,  the  amount  of  water 
that  can  be  pumped  in  1  hour  will  be  16.32  X  60  =  979.2  gal. 

Ans. 

(15)  The   amount   of  water  to  be  delivered   per  minute 
will  be  -9T3g®  =  15.6  gallons.     The  velocity  of  flow  for  this  case 
should  not  exceed  500  feet  per  minute.     Applying  rule  4, 
we  have 

.       19.25  X  15.6 

A  = r^r      —  =  .601  square  inch,  nearly. 

500 

From  Table  VI  we  see  that  the  nearest  larger  size  pipe  is 
the  1-inch  pipe,  which  has  an  actual  area  of  .863  square 
inch.  Hence,  a  1-inch  pipe  should  be  used. 

(16)  Bends  in  a  pipe  greatly  increase  the  resistance  to 
the  flow  of  water,  and,   hence,  also  the  power  required  to 
move  the  water  through  the  pipe.     As  the  power  available 
for  the  suction  pipe  of  a  pump  is  quite  small,  any  increase 
in  the  resistance  to  the   flow  will   materially  interfere  with 
trie  satisfactory  working  of  the  pump.     See  Art.  39. 

(17)  See  Art.  41. 

(18)  About  30  in.,  as  explained  in  Art.  43. 


§  6  MECHANICS  OF  FLUIDS.  3 

(19)  By  means  of  a  vacuum  gauge.     See  Arts.  45  to  47, 
inclusive. 

(20)  Applying  the  principle  of  Art.  48,  we  have  - 
—  163.2  in.     Ans. 

(21)  The  pressure  of  the  atmosphere  is  measured  with 
an  instrument  called  a    barometer.     See  Arts.   49  to  52, 
inclusive. 

(22)  The  higher  the  elevation  the  shorter  is  the  column 
of  air  and  the  lower  the  density  of  the  air.     Consequently, 
the  pressure  exerted  by  the  air  becomes  less  the  higher  we 
go  above  sea  level.     See  Art.  53. 

(23)  Like  the  pressure  exerted  by  a  liquid,  the  pressure 
of  the  atmosphere  acts  in  all  directions  with  the  same  inten- 
sity.    See  Art.  54. 

(24)  The  pressure  of  a  gas  varies  inversely  as  the  volume, 
the    temperature    remaining    constant.     Thus,    if   the    vol- 
ume  is   reduced  one-half,    the    pressure    is  doubled;   if  the 
volume  is   reduced    to  one-third    the    original  volume,   the 
pressure  is  increased  to  three  times  the  original  pressure. 
See  Art.  55. 

(25)  (a)  Gauge  pressure  is  pressure  measured  above  the 
atmospheric  pressure.     See  Art.  56. 

(b]  Absolute  pressure  is  pressure  measured  from  vacuum 
and  is  equal  to  gauge  pressure  plus  atmospheric  pressure. 
See  Art.  56. 

(26)  Call  the  original  volume,  or  volume  of  the  tube,  1  ; 
then  the  volume  after  the  piston  has  moved  |  the  length  of 
the  tube  will  be  1  —  f  =  ^.     Applying  rule  7,  we  have 

14.7  x  1 


,  j 

(27)     Applying  rule  8, 
14.7  X  1 


—  73.5  Ib.  per  sq.  in.     Ans 


147 


=  .1  =  ytg.  the  original  volume.     Ans. 


4  MECHANICS  OF  FLUIDS.  §  6 

(28)  An  air  pump  is  an  apparatus  for  removing  air  from 
any  vessel ;  in  other  words,  it  is  an  apparatus  for  producing 
a  vacuum  in  any  vessel.      See  Art.  6O. 

(29)  No.      See  Art.  61. 

(30)  See  Art.  62. 

(31)  See  Arts.  63  and  64. 

(32)  When  the  water  reaches  the  same  level  in  both  ves- 
sels,  the  flow  of  water  from  one  vessel  into  the  other  will 
cease.     This  follows  from  Arts.  63  and  64. 

(33)  The  highest  point  of  the  siphon  should  not  be  more 
than  28  feet  above   the  level  of  the  water    that    is    being 
siphoned,  or  the  siphon  will  not  work.      See  Art.  65. 

(34)  (a)  A  pump  is  an  apparatus  for  raising  water.      See 
Art.  68. 

(b]  Pumps  may  be  divided  into  three  classes,  depending 
on  their  mode  of  operation.      See  Art.  69. 

(c)  (1)   Suction  pumps;  (2)  lifting  pumps;  and  (3)  force 
pumps.     See  Art.  69. 

(35)  A  suction  pump  acts  by  producing  a  vacuum  in  the 
suction  pipe,  which  causes  the  water  into  which  the  lower 
end  of  the  suction  pipe  is  immersed  to  rise  in  the  suction 
pipe.     See  Art.  7O. 

(36)  A  lifting  pump  will  raise  water  to  a  higher  point 
than  a  suction  pump.     See  Art.  72. 

(37)  A  force  pump  piston  does  not  have  any  valve  like 
the  piston  of  a  lifting  pump.      See  Art.  73. 

(38)  A   single-acting   force   pump    discharges    water    at 
every  second  stroke ;  a  double-acting  force  pump  discharges 
water  during  each  stroke.     See  Art.  76. 


STRENGTH  OF  MATERIALS. 


(1)  (a)  A  stress  is  a  force  that  acts  in  the  interior  of  a 
body  and  tends  to  produce  a  deformation  in  the  body.     See 
Art.  1. 

(b}  Tensile,  compressive,  shearing,  transverse,  and  tor- 
sional.  See  Art.  2. 

(2)  By  Art.  3,  we  have 

unit  stress  =  ££fo  —  400  lb.  per  sq.  in.     Ans. 

(3)  (a)  Strain  is  the  amount  of  deformation  produced  in 
a  body  by  the  action  of  a  stress.     See  Art.  4. 

(b}  Elasticity  is  that  property  of  a  body  by  virtue  of 
which  it  will  tend  to  return  to  its  original  shape  after  the 
force  producing  the  deformation  is  removed.  See  Art.  5. 

(c]  The  elastic  limit  is  that  unit  stress  at  which  permanent 
set  begins.  See  Art.  7. 

(4)  Since  the  ultimate  tensile  strength  per  square  inch  is 
55,000  pounds,  the  force  at  which  a  bar  having  an  area  of 
4  square  inches  will  rupture  is  4  X  55,000  =  220,000  lb.    Ans. 
See  Art.  1O. 

(5)  Annealing  makes  the  chain   tougher,  more  ductile, 
and  less  liable  to  break  from  sudden  jerks.     See  Art.  16. 

(6)  Untwist  a  portion  of  the  rope  and  examine  the  con 
dition  of  the  inner  surfaces.     See  Art.  2O. 

§? 


2  STRENGTH  OF  MATERIALS.  §  7 

(7)  (a)  The  safe  load  to  be  lifted  depends  on  the  size  of 
the  rope  and  on  its  manner  of  attachment  to  the  hook  of  the 
tackle  block  or  to  the  work.     See  Art.  23. 

(b)     See  Art.  23. 

(8)  (a)   Manila  rope  is  chiefly  used  for  hoisting  and  for 
power  transmission.     See  Art.  2G.         • 

(b}     The   fibers  are  oiled  to  reduce  internal   wear.      See 
Art.  26. 

(9)  The  diameter  of  the  pulley  should  be  at  least  40  times 
the  diameter  of  the  rope.      See  Art.   28. 

(10)  Applying  rule  4,  Art.  3O, 

W=  600  X  (1|)*  =  1,350  Ib.     Ans. 

(11)  Applying  rule  0,  Art.  3O, 

C=  |/16,000  X  .0316  =  4  in.,  nearly.     Ans. 

(12)  A  column  both  of  whose  ends  are  fixed  is  4  times  as 
strong  as  one  that  has  both  ends  movable  and   1^  times  as 
strong  as  one  that  has  one  fixed  end  and  one   movable  end. 
See  Art.  34. 

(13)  See  Art.  34. 

(14)  Applying  rule    7,  Art.   37,  and  taking  the  proper 
values  from  Tables  V  and  VII,  we  have 

T/r/      14,000  X  .7854  X  14' 

W=  — —  —  =  1,291,480  Ib.     Ans. 

(16  X  12)' 

"^  281. 25  X  14' 

(15)  By   rule   8,  Art.   42,   the  area    of   the    piston    rod 
should  be 

.7854  X  40'  X  110 

—  =  38.397  square  inches. 

Oj  uOO 

/3S.397 

The  corresponding  diameter  ISA  /  =  7  inches,  nearly. 

y     .7854 

Hence  a  6"  piston  rod  is  too  small.     Ans. 


§  7  STRENGTH  OF  MATERIALS.  3 

(16)  From  Table  X,  5  for  this  case  is  found  to  be  1,460. 
From  Table  XI,  1?  —  1    X  U  =  457.33.     Now,  applying  the 
proper  formula  from  Table  IX,  we  have 

„,      4X1,460X457.33 

12  X  12  =  1*,MT  Ib.     Ans. 

(17)  By  rule  9,  Art.  5O,  and  Table  XII,  the  safe  load 
would  be  4  X  6  x  500  =  12,000  pounds.      Hence,  a  load  of 
10,000  pounds  is  safe.     Ans. 

(18)  From   Table   XII,  5  —  4,400.      Applying  rule  1O, 
Art.  51,  we  have 

400,000 

a  =  4,400  ~~  :    sq-  m>  '  nearly-  Ans- 

(19)  Double  shear  means  that  the  body  subjected  to  the 
shearing  stress  resists  this  stress  at  two  planes,  at  both  of 
which  shear  must  occur  to  produce  failure.     See  Art.  48. 

(20)  Countershafts  serve  to  effect  changes  in  speed  and  to 
stop  and  start  the  machinery.     See  Art.  54. 

(21)  Cold-rolled  shafting  is  made  cylindrically  true  by  a 
special  rolling  process.     Bright  shafting  is  turned  up  in  a 
lathe.     See  Art.  55. 

(22)  The  diameter  by  which  bright  shafting  is  designated 
is  that  of  the  bar  from  which  it  is  turned.     See  Art.  56. 

(23)  Pulleys  should  be  placed  as  near  the  bearings  as  pos- 
sible so  that  the  deflection  of  the  shaft  may  be  as  small  as 
possible.     See  Art.  58. 

(24)  Applying    rule  11,  Art.    6O,    after    finding    from 
Table  XIV  that  C  for  this  case  is  85,  we  have 

H  =  88*8°  =  481.8,  nearly,  say  482  H.  P.     Ans. 
oo 

(25)  From  Table  XIV  we  find  C  for  this  case  to  be  65. 
Applying  rule  IS,  Art.  61, 


H.     S.     I.—31 


4  STRENGTH  OF  MATERIALS.  §  7 

(26)     From    Table    XIV,    C  =  95.       Applying    rule    13, 
Art.  62, 

=  -P§0,sa3in.     Ans. 


(27)  The  horsepower  of  a  shaft  will  vary  directly  with 
the  speed.     See  Art.  63. 

(28)  See  Art.  11. 

(29)  A  steel  rope  wears  better  than  an  iron  rope.     See 
Art.  29, 


INDEX. 


NOTE. — All  items  in  this  index  refer  first  to  the  section  (see  the  Preface)  and 
then  to  the  page  of  the  section.  Thus,  "Acceleration  47"  means  that  acceleration 
will  be  found  on  page  7  of  section  4. 


A.                  Sec. 
Absolute  and  gauge  pressure.  .      6 

Page. 
37 
1 

Arc  of  contact  of  belts,  To  find 
the 

Sec. 
5 

Page. 

14 

15 

Area  

H 

18 

"            force,  Constant.  ..      4 
Acceleration  4 

16 

7 

Arithmetic,  Definition  of  
Atmosphere,  Pressure  of  

1 
6 

4 

1 

28 
1 

gravity  4 

16 

Average  velocity  

4 

7 

Work  of  4 

24 
31 

Avoirdupois  weight  
Axle 

2 
5 

11 
1 

"          circle                               5 

31 

5 

5 

"          line  5 

31 

Addition  1 
"         of  decimals  1 

4 
39 

B. 

Backlash  

Sec. 
5 

Page. 
31 

5 

10 

bers  2 
"         of  fractions  1 

16 
28 
14 

Standing  
Balancing  pulleys  

5 

5 
g 

9 
9 
32 

2 

fi 

33 

Aggregation,  Signs  of.  1 

52 
40 

"           Mercurial  
Base   Definition  of 

6 

<t 

32 

17 

2 

3 

37 

4 

38 

Altitude,  Definition  of  3 
"         of  cone  3 

17 
36 

Bearings,  Allowable  pressures 
on  

4 

37 

23 

Beaume's  hydrometer  

e 

21 

Amount  in  percentage  2 

3 
33 

Belt,  To  find   the  horsepower 
of  a        

5 

16 

Angle  3 

14 
27 

<;      To  find  the  length  of  a..  . 
Belts 

5 

5 

13 

12 

"     Vertex  of  3 
Angles  or  arcs,  Measure  of  2 

14 

12 

"      Calculations  for  
"      Care  and  use  of  

5 

5 

13 
19 

38 

5 

23 

50 

"      Cotton  

5 

13 

Archimedes,  Principle  of  6 

19 

"      Double..,. 

5 

18 

xii 


Sec. 

Page. 

Sec. 

Page. 

Belts,  Effective  pull  of.  5 

14 

Chain  hoist,  Differential  5 

47 

"      Flapping  of  5 

20 

Chains  7 

6 

"      Joining  the  ends  of  5 

21 

"       Strength  of.  7 

7 

"      Lacing  5 

21 

Change  in  motion  of  a  body  —      4 

17 

13 

Circle  3 

14 

"      Rubber  5 

13 

"      Addendum  5 

31 

"      Rubber,  Care  of  5 

24 

"      Definition  of  3 

27 

'      Speedof  5 

16 

"      Pitch  5 

29 

"     Stretch  of  5 

24 

"      Root  5 

31 

"     To  find  the  arc  of  contact 

Circular  pitch  5 

30 

of  5 

14 

"         pitch  system  5 

34 

"      To  find  the  width  of  5 

14 

Circumference  3 

14 

Bevel  gear  5 

28 

Classification  of  pumps  6 

48 

Black  shafting  7 

33 

"            of  stresses  7 

1 

Blackwall  hitch  7 

11 

Coach  screws  5 

59 

Block  5 

42 

Coefficient  of  friction.  4 

33 

"      Weston  differential  pul- 

Coefficients of  friction  4 

35 

ley  5 

46 

Cold-rolled  shafting  7 

32 

Blow,  Force  of  a  4 

25 

Columns.  Strength  of  7 

16 

Body,  Change  in  motion  of  a..      4 

17 

Combination  of  pulleys  5 

43 

Bodies  4 

2 

"             of  pulleys,    Law 

"      Aeriform  4 

2 

of  5 

44 

"      Gaseous  4 

2 

Common  denominator  1 

26 

"      Liquid  4 

2 

Comparison  of  forces  4 

10 

"      Solid  4 

2 

Components  of  a  force  4 

29 

Bolts  5 

57 

Composition  of  forces  4 

26 

"     Machine  5 

57 

Compound    denominate   num- 

Brace        1 

52 

ber  2 

10 

"     3 

2 

lever  5 

3 

Brackets  1 

52 

"            lever,  Law  of  5 

5 

2 

"            numbers  2 

9 

Bright  shafting  7 

32 

Compressibility  4 

4 

Brittleness  4 

5 

1 

Buoyant  effects  of  water  6 

19 

Cone  3 

36 

"      Frustum  of  3 

37 

C.                    Sec. 

Page. 

Conservation  of  energy  4 

24 

Cable-laid  or  hawser-laid  rope.      7 

7 

Constant  accelerating  force.  ...      4 

16 

Calculations  for  belts  5 

13 

"         retarding  force  4 

16 

"             involving  percent- 

Constants for  cast-iron  pillars.      7 

19 

age  2 

4 

"          for  line  shafting  7 

35 

"            relating  to  gears..      5 

34 

"          for  wooden  pillars..      7 

20 

Cancelation  1 

20 

"         for    w  r  o  u  g  h  t-iron 

Capacity,  Measures  of  2 

12 

and  structural- 

Capscrews  5 

57 

steel  pillars  7 

18 

Care  and  use  of  belts  5 

19 

Convex  surface  3 

33 

"    of  rubber  belts  5 

24 

Cotton  belts  5 

13 

Cast-iron  pillars,  Constants  for      7 

19 

Crowning  pulley  faces  5 

8 

Cementing  belts  5 

23 

Crushing  strength  of  materials      7 

13 

Center  3 

14 

Cube  3 

32 

"        Definition  of  3 

27 

"      root  2 

35 

"        of  gravity  4 

15 

11 

"       of  gravity  4 

39 

Curve,  Epicycloidal  5 

33 

"        of  gravity  of  a  solid  4 

42 

"       Hypocycloidal  5 

33 

Centrifugal  force  4 

43 

"       Involute  5 

33 

Centripetal  force  4 

44 

Curved  line  3 

13 

Chain  hoist  5 

46 

Cycloidal  tooth  5 

32 

INDEX. 


Cylinder 

&'C. 

y 

Page. 
32 

Divisor            

S,-c. 
1 

Page. 
16 

Cylinders   Pitch 

5 

29 

6 

54 

«t 

40 

"       belts             

5 

18 

"           ring,  Convex  area 

Downward  pressure  

6 

3 

"          ring,  Volume  of... 
D. 

3 

3 

Ste. 

6 

40 
41 

Page. 
42 

"           pressure,    General 
law  of  
"           pressure  of  a  fluid. 
Driven  wheel  
Driver                                      .... 

6 
6 
5 
5 

5 

3 
25 
10 

1 

37 

5 

25 

5 

25 

fraction  

1 

50 

Drv  measure  

f 

10 

Decimals,  Addition  of  
"           Definition  of  
"           Division  of  
"           Multiplication  of  .. 
"           Reading  of 

1 

1 
1 
1 
1 

39 
37 
44 
42 
37 

Ductility  
Dynamics  

E. 

4 
4 

Sec. 
T 

5 

15 

Page. 
32 

"           Subtraction  of  

1 

41 

Effective  pull  of  belts  

5 

14 

Definition  of  mechanics  

4 

1 

Effects  of  lateral  pressure  of  a 

3 

15 

fluid  

6 

8 

"        and  limits  of  exhaus- 
tion               

ff 

42 

Efficiency  
Elastic  limit  

5 

7 

60 
2 

* 

g 

Elasticity                 

4 

4 

2 

o 

16 

^ 

2 

"           numbers,  Division 
of  
"            numbers,  Multipli- 
cation of  

2 
2 

21 
20 

Energy  
"        Conservation  of  
"        Kinetic  
Potential  

4 
4 

4 
4 
5 

22 
24 
22 
23 
32 

0 

13 

5 

33 

"            numbers,  Subtrac- 
tion of  

9 

18 

"             tooth  

5 

32 

1 

of  

s 

20 

s 

27 

Equilibrium         ...            

4 

45 

Pitch  

5 
5 

29 
31 

"            Neutral  
"            Stable 

4 
4 

46 
45 

5 

35 

"             Unstable 

4 

46 

2 

3 

2 

27 

Differential  chain  hoist  
"            pulley  block,  Wes- 

5 

47 

Exhaustion,  Degrees  and  limits 
of           

6 

42 

5 

46 

Expansibility     

4 

4 

Dimensions  of  standard  pipe.  . 

6 

24 

Extension  

4 

3 

wrought-iron 

6 

25 

F. 

Faces  of  a  solid  

v,  e. 
3 

Page. 
32 

Direct  proportion  
Distance     between    line-shaft 
bearings  

2 

7 
1 

44 

33 
16 

Factors  of  a  number  

Fastenings,  Screws  used  as.  ... 
Fixed  and  movable  pulleys.... 

1 

5 
5 
5 

20 
56 
42 
42 

Divisibility 

4 

3 

5 

20 

Division  

16 

Flow  of  water  in  pipes  

6 

22 

•     "       of  decimals  
"        of    denominate    num- 
bers   
"        of  fractions..  .. 

1 

2 
1 

44 

21 
33 

Fluid,  Downward  pressure  of  a 
"      Effects  of  lateral  pres- 
sure of  a  
"      Lateral  pressure  of.  ... 

6 

0 
I 

3 

8 
6 

xiv 


Sec.  Page. 

Fluid,  Upward  pressure  of  a.. .  6  5 

Fluids,  Mechanics  of 6  1 

Follower  and  driver 5  25 

Foot-pound 4  20 

Force 4  10 

"      Accelerating 4  15 

"      arm 5  2 

"      Centrifugal 4  43 

"      Centripetal 4  44 

"      Components  of  a 4  29 

"      Constant  accelerating. . .  4  16 

"      Constant  retarding 4  1C 

"      General  principles  of.. ..  4  10 

"      of  a  blow 4  25 

"      of  gravity 4  15 

"      of  gravity,  Acceleration 

due  to 4  16 

"      pump,  Double-acting 6  54 

"      pumps 6  51 

"      Retarding 4  15 

"      Time  effect  of  a 4  19 

Forces,  Comparison  of 4  10 

"        Composition  of 4  26 

"        Names  of 4  10 

Reduction  of 4  11 

"        Resolution  of 4  29 

"       Triangle  of 4  26 

Forms  of  gear-teeth 5  32 

"       of  keys 5  52 

Formula,  Definition  of 3  1 

Fraction,  Reduction  of  a,  to  a 

decimal 

Fractions 

"          Addition  of 

"          Division  of 

"          Multiplication  of 31 

u          Reduction  of 25 

"         Subtraction  of 

Friction 

"       Coefficient  of 

"        Coefficients  of 

"        Laws  of 

"        of  a  screw 53 

Frictional  losses  with  pulleys.  44 

Frustum  of  cone 37 

"         of  pyramid 37 

Q.                    Sec.  Page. 

Gas,  Permanent 4  2 

Gases,  Pressure  of 6  27 

"       Specific    gravity  and 

weight  per  cubic  footof  6  17 

"      Tension  of 6  36 

Gaseous  bodies 4  2 

Gauge,  Pressure 6  31 

"       Vacuum 6  31 


Gear 

"      Bevel 

"      calculations..... 

"      Miter 

"      Spur 

"      teeth,  Forms  of 

"      teeth,  Proportions  of 

"      wheels 

"      wheels,  Teeth  of 

Gears,  Number  and  speed  of 

teeth  of  

General  properties  of  matter.. 

Globe 

Gravity,   Acceleration   due   to 

the  force  of 

"          Center  of 

"          Center  of 

"          Force  of 

11         Influence  of,  on 

liquids 

"          Specific 

Gravitation 

"          Law  of 

H. 

Hardness 

Hemp  ropes 

Heptagon 

He  xagon 

Horizontal  line 

Horsepower    

of  a  belt 

Hydraulic  jack,   Watson-Still- 


Hydrokinetics 

Hydrometer 

"  Beaume's 

Nicholson's 

"  of  constant  weight 
Hydrometers  of  constant  vol- 
umes   

Hydrostatic  pressure 

Hydrostatics 

Hypocycloid 

Hypocycloidal  curve 

Hypotenuse 


Page. 

27 
28 
34 
28 
27 
32 
31 
27 
29 


15 
15 
17 

Page. 
5 


Sec.  Page. 


Impenetrability 4 

Improper  fraction 1 

Inclined  plane 5 

"         plane,  Applications  of 

the 5 

Indestructibility 4 

Inertia ...  4 


XV 


Sec. 
4 

Page. 
12 

Sec. 
5 

Page. 
31 

4 

12 

T 

13 

4 

12 

"      Pitch 

5 

29 

4 

12 

"      Root 

5 

31 

Inverse  proportion  
"       proportion  

2 

9, 

44 
46 

"      shaft  bearings,  Distance 
between  

33 

"       ratio  

9, 

41 

7 

32 

Involute  
"         curve  
"        teeth                        .  . 

5 
5 

5 

32 
33 
33 

"      shafting,  Constants  for  .  . 
"      shafting.   Rules  and  for- 

35 
35 

Involution  
Isosceles  triangle,  Definition  of 

2 
3 

25 
20 

"      Vertical  
Linear  measure  

3 

4 

13 
10 
2 

j 

Sec 

Page 

f 

10 

5 

21 

8 

12 

K. 

Keys,  Forms  of  

Sec. 
5 
5 

Page. 
53 
50 

Liquids  influenced  by  gravity. 
Specific    gravity    and 
weight    per     cubic 

6 

8 
16 

Kinetic  energy  

4 

22 

Location  of  the  resultant  

4~ 

28 

4 

15 

7 

15 

L. 

y,.,- 

Page 

"      columns,  Safe  working 

7 

17 

Lacing  belts  

5 

21 

8 

11 

Lagscrews  
Lateral  pressure.  General  law 
of 

5 
6 

59 

M.                 i 

•tec. 
5 

Page. 
57 

6 

| 

5 

1 

"       pressure  of  a  fluid,  Ef- 

"       screws  

5 

57 

fects  of  
Law  for  downward  pressure  of 
a  fluid 

6 
6 

8 
5 

Malleability  
Manila  rope  

4 
7 
6 

5 

12 
37 

"    for  lateral   pressure  of  a 
fluid 

6 

7 

"          law,  Application  of. 
Mass 

4 

38 

ir 

"    for  upward  pressure  of  a 
fluid  
"    of  combination  of  pulleys 
"    of  compound  lever  

6 
5 
5 
4 

6 
44 
5 

17 

Materials,  Shearing  strength  of 
"         Tensile  strength  of.. 
"         Transverse  strength 
of  
Matter                        

7 

7 

4 

30 
2 

23 
1 

4 

12 

4 

1 

"    of  Mariotte 

6 

37 

4 

3 

"    ofPascal  

6 

3 

"        Special  properties  of.  .  . 

4 

3 

5 

2 

4 

7 

Laws  of  friction  
"      of  motion,  Newton's  
"     of  the  screw  

4 
4 
5 
4 

11 
54 
17 

Measure,  Cubic  
"         Dry  
"         Linear  
"         Liquid  

2 
2 
2 
? 

11 
12 
10 
12 

Least  common  denominator  

1 
5 

26 
13 

"         of  angles  or  arcs  

2 
fl 

12 
13 

5 

13 

<> 

12 

5 

3 

o 

10 

5 

5 

<?, 

10 

"       Law  of  the  

5 
5 

2 

5 

'•         of  capacity  

2 
9 

12 
10 

6 

49 

"         of  weight    

9 

11 

Like  numbers.  .  .  . 

1 

1 

Measurement  of  vacuum  

6 

29 

INDEX. 


Mechanics,  Definition  of  
offluids  
"           Principles  of  
Mensuration,  Definition  of  
Mercurial  barometer  

Sec.  Page. 
4          1 
6          1 
4          1 
3        13 
6        32 
5        56 

6        15 
3        15 
5        28 
1        24 
4          3 
4          1 
4        19 
2        13 
4          5 
4        17 
4          5 
4        11 
4          6 
5        42 
5        43 
1        11 
1        11 
1        42 

2        20 
1        31 
1        11 

Sec.  Page. 

4        10 

.5 
Percentage,    Calculations    in- 
volving   
"             Definition  of  
Permanent  gas  

'ec.  Page. 

2         4 
2         1 
4         2 
3        13 
5        28 
6        25 
6        22 
5        29 
5        30 
5        34 
5        29 
5        29 
5        31 
5        35 
5        29 
5        53 
7         7 

5        50 
3        17 
5        47 
6        51 
6        40 
6        27 
3       24 
3        24 
3        24 
4          3 
4        23 
4        10 
4        21 

6        34 
6          3 

6         3 

6        31 
6          1 
6          6 
6        27 
6        28 
4        30 
6          5 
6          5 
6        37 

4       37 
1        20 
1        20 
6        19 
4          1 
3        32 
3        32 
4        38 
1        24 

Pinion 

Metals,    Specific    gravity    and 
weight  per  cubic  foot  of  

Pipes,  Flow  of  water  in  

"     Circular  
"     Circular,  system  
"     cylinders  

Mixed  number  
Mobility...  

"     Diametral  
"      Diametral,  system  
"     line                        

"        Change  in  the,  of  a  body 

"     of  a  screw  
Plain-laid  rope  

"        Newton's  laws  of  
"        Path  of  a  body  in  
Movable  and  fixed  pulleys  

Plane,  Applications  of  the  in- 
clined    
"      figure,  Definition  of  a.  .. 
"      Inclined  

Multiplicand 

Multiplication  
"              of  decimals  
"              of     denominate 
numbers  
"              of  fractions  
Multiplier 

Pneumatics  
Polygon,  Definition  of  a  
"          Regular  
Polygons  

N. 

Neutral  equilibrium  
Newton's  laws  of  motion  
Nicholson's  hvdrometer  

4 
4 
fi 

46 
11 
21 
1 
1 

38 
9 
1 
23 

Page. 
14 

Page. 
13 
17 
32 
52 
2 
29 
3 
6 
24 

Pressure    at    different    eleva- 
tions, Variations  of 
Downward  

Notation  
Number,  Definition  of  
"         of  teeth  and  speed  of 
gears  
Numbers,  Denominate  

1 

1 

5 
2 
1 
1 

Sec. 
3 

Sec. 
3 
3 
3 
1 
3 
6 
6 
4 
3 

fluid  
gauge  
"           Hydrostatic  
"           Lateral,  of  a  fluid... 

"           of  the  atmosphere.  . 
"           Tangential  
"            Upward  

0. 

Obtuse  angle  

P. 

"           Upward,  of  a  fluid.. 
Pressures,  Absolute  and  gauge 
"           Allowable,  on  bear- 
ings   
Prime  factor  

Parallelogram,  Definition  of.  .  . 

Parenthesis  . 

Principle  of  Archimedes  
Principles  of  mechanics  

Partial  vacuum  
Pascal's  law  
Path  of  a  body  in  motion  
Pentagon  ... 

Projected  area  of  bearing  
Proper  fraction  .... 

xvii 


Properties  of  matter  
Proportion  

Sff. 

4 
2 
2 
2 

5 
5 

5 
5 
5 
5 
5 
5 
5 
5 
5 

Page. 
\ 
43 
44 
44 
31 
14 

46 
8 
42 

43 

7 
7 
9 
43 
42 

Resistance  of  inertia  
Resolution  of  forces  
Resultant  
"          Location  of  the  
Retarding  force  
"          force,  Constant  
Retardation  
"            Work  of  
Rhomboid,  Definition  of  
Rhombus,  Definition  of  
Right  angle  
"     angled  triangle,   Defini- 
tion of                         .... 

Stf. 
4 

4 
4 
4 
4 
4 
4 
4 
3 
3 
3 

3 
3 
5 
5 
2 
5 
2 
7 
7 
7 
7 

7 
7 
7 
5 
5 

7 

7 
7 
5 

7 
4 
5 
5 

Sec. 

7 
3 
5 
5 
5 
5 
5 
5 
5 
5 
5 
5 
3 

Page. 
12 
29 
27 
28 
15 
16 
7 
24 
17 
17 
14 

20 
20 
31 
31 
35 
31 
28 
7 
12 
7 
7 

12 
13 
7 
13 
24 

35 

5 
29 
10 

13 
7 
25 
10 

Page. 

17 
20 
53 
58 
54 
53 
59 
57 
56 
56 
56 
59 
15 

Direct  
"           Inverse  
Proportions  of  gear-teeth  
Pull  of  belts,  Effective  
Pulley  block,  Weston  differen- 
tial   
"      Crowning  faces  of  
•'       Fixed  

"       Movable  
"      Split  
Pulleys  

"        Balancing  
"        Combination  of  
"        Fixed  and  movable.  .. 

"     triangle  
Root  

"     circle  

"       Frictional  losses  with. 
"        Law  of  combination  of 
"        Rules  for  speeds  of.... 
Solid  
Pump,  Lifting  

5 
5 
5 
5 
6 
6 
6 
6 
6 
6 
3 
3 

Sec. 
3 
3 
3 

1 

Sec. 
5 

44 
44 

10 
7 
49 
48 
47 
48 
51 
51 
36 
37 

Page. 

17 

1 
16 

Page. 
29 

"     Cube  
"     line  
"     Square  
Rope,  Cable-laid  or  hawser-laid 
"     Manila 

"     Plain-laid 

Classification  of  
"        Force  

"     Strength  of  manila  hoist- 
ing  
"      Wire  
Ropes,  Strength  of  
Rubber  belts 

"        Plunger  

Pyramid  
"        Frustum  of  

Q. 

Quadrilateral,  Definition  of.  .  .  . 
Quadrilaterals  
Quantitv  

"        belts,  Care  of  
Rules  and    formulas    for   line 
shafting  
"      and  formulas  for  tensile 
strength  
"      for  shearing  
"      for  speeds  of  pulleys  
"      for  the  strength  of  wire 

Quotient  

R. 

Rack  

Rate  per  cent  

2 
2 

2 
2 
2 
3 

1 

1 
1 

2 
4 
1 
3 
4 

3 
40 
41 
41 
41 

17 

50 
48 
49 

13 
11 
25 
24 

12 

"      for  wheel  work  
Running  balance  

S. 

Safe  working  load  on  long  col- 

Ratio  

"     Inverse  
"     Reciprocal  
Reciprocal  ratio  
Rectangle,  Definition  of  
Reducing  a  decimal  to  a  frac- 
tion . 

Scalene  triangle,  Definition  of. 
Screw 

"          a  fraction  to  a  deci- 

"          inches  to   a  decimal 
part  of  a  foot  
Reduction  of  denominate  num- 
bers   
"           of  forces  
"           of  fractions  
Regular  polygon  
Resistance  due  to  inertia.... 

"       Pitch  of  a  

Screws,  Coach  
Machine  
Metal  
"       used  as  fastenings  .... 
"        Wood 

"        Wood 

Seconds  .... 

1 

>ec. 
g 

Page. 
28 

See. 

2 

Page. 
37 

3 

28 

4 

45 

Shafting,  Black  
"           Bright  
"          Cold-rolled  

32 
32 

Standard  of  weight  
"         pipe  dimensions  
Standing  balance  

4 
6 

5 

3 
24 
9 

Statics 

4 

15 

^ 

29 

7 

•'          Rules  for 

„ 

29 

„ 

"          strength  of  materials 
Short  column  
"     methods   of   abstracting 

7 
o 

30 
15 

34 

Strength  of  chains  
"         of  columns  . 
"         of  manila  ho  i  s  ting1 

7 
7 

16 

12 

7 

9 

7 

7 

7 

29 

5 

31 

52 

Stress 

7 

1 

Simple  denominate  numbers.. 
Siphon,  Theory  of  the  
Slant  height  of  cone  or  pyra- 

2 

6 

T 

9 
44 

36 

"      Unitof  
Stresses,  Classification  of  
Stretch  of  belts  
Subtraction  

7 

5 
1 

1 

1 
24 
9 

Slings                       

7 

9 

1 

41 

Use  of  

10 

"           of  denominate 

Solid  bodies  
"      Center  of  gravity  of  a  
"      Edges  of  a..   
"      Faces  of  a   

4 
4 
3 
3 
5 

2 

42 
32 
32 

7 

numbers  
"           of  fractions  
Suction  pump  

T. 

2 
1 
6 

Set 

18 
29 
48 

Page 

Special  properties  of  matter... 
Specific  gravity  

4 
6 

3 
15 

Tangential  pressure  
Taper  of  keys  

4 

5 

30 
50 

5 

33 

cubic  foot  of  gases  .  . 
"       gravity  and  weight  per 
cubic  foot  of  liquids. 
"       gravity  and  weightper 

6 
6 

17 
16 

"      of  gear-wheels  
Tenacity  
Tensile  strength  of  materials.. 
"        strength,   Rules  and 

5 
4 
7 

29 
5 
2 

5 

ff 

15 

$ 

36 

1 

24 

fi 

44 

substances  

6 

18 

Thread  

>i 

53 

"        gravity  and  weightper 
cubic  foot  of  woods. 

6 

16 

"       of  a  screw  

5 
4 

53 
19 

Speed  and  number  of  teeth  of 

2 

12 

gears  
"      of  belts       ...            

5 

5 

38 
16 

To  find  the  arc  of  contact  of 
belts                                 ..  . 

5 

14 

Speeds  of  pulleys,  Rules  for.  .  . 

5 

1 

10 
39 

"  find  the  length  of  a  belt  

5 
5 

13 
14 

g 

5 

32 

Split  pulley 

5 

5 

32 

Spur  gear  

5 

27 

7 

32 

Square,  Definition  of  
foot  

3 
fl 

17 
18 

Transverse  strength  of  mate- 

7 

23 

inch  
"       measure  

3 
2 

18 
10 
28 

Trapezoid,  Definition  of  
Triangle,  Definition  of  

3 
3 
3 

20 
20 

"       root.  Short  method  of 

"         Isosceles  

3 

20 

working  

2 

34 

"         offerees  

4 

26 

Sec. 
Triangle,  Right-angle  3 
"          Scalene  3 
Troyweight  2 

U.                  Sec. 

INDEX. 
Page. 

20              Vinrnlnm 

See. 
3 
3 
3 
3 

Sec. 

6 
4 
4 
5 
2 
4 
2 
4 
2 

5 
5 
5 
5 
5 
5 
5 

7 
5 
5 

7 

6 
4 

4 
4 
4 
5 
6 

6 

Stt. 

7 

XIX 

Page. 
2 
33 
41 
33 

Page. 

11 
3 
15 

11 
17 
11 
3 
11 

46 
1 
5 
25 
25 
25 
14 
13 

13 
56 
59 
20 

16 
20 

24 
20 
20 
29 

29 

18 

25 

Page 
7 

20 
11 

Page 
6 
6 
1 
48 
33 
20 
18 
13 
-  1 
46 
5 

6 
5 
19 
10 

Page. 
29 
31 
29 
24 
2 
6 

34 

18 
6 
6 

Volume  

"        of  cylindrical  ring.... 
"        Unit  of  

W. 

Watson-Stillman  hydraulic 
jack  

Uniformly  varying  velocity.  .. 
Unit,  Definition  of  
"      method  

Weight 

"      of  work   

"       arm  
"       Avoirdupois..  
"        Lawsof  
"        Measures  of  
"        Standard  of 

"      square  
United  States  money  

"        pressure,  General  law 
of                                        6 

Weston  differential     pulley 
block 

"         pressure  of  a  fluid  6 
Use  and  care  of  belts  5 
"    of  slings  7 

V.                  Sec. 
Vacuum  6 
"         gauge  6 

"     and  axle  

"     Driven  

"      work  
"     work,  Rules  for  
Width  of  belts,  To  find  the  .... 
Wire  rope  
"      rope,  Rules  for  strength 
of  

"         Measurement  of  6 
Value  of  a  fraction  1 
Vapor  4 

Variations  of  pressures  at  dif- 

Wooden  pillars,  Constants  for. 
Woods,   Specific    gravity    and 
weight  per  cubic  foot  of  
Work 

Various    substances,    Specific 
gravity  and  weight  per  cubic 
foot  of  6 
Varying  velocity,  Uniformly.  .      4 
Velocity  

'•     of  acceleration   and  re- 
tardation   

"      power,  and  energy  
"      Unit  of  

"         Average  

"         Mean  
"         problems,  Rules  for.. 
"         ratio  

7 

59 
6 
6 
6 
14 
36 
36 
13 
52 

Worm                       

"      wheel  

Wrought-iron  and  structural- 
steel    pillars,    Con- 

"         Uniformly  varying..  . 
"         Variable  

"         iron  welded    pipes, 
Standard   dimen- 
sions of  

Y. 

Yarn... 

Vertex  of  angle  3 
"        of  cone  3 
"        of  pyramid  3 
Vertical  Hue  3 

Vinculum  ,      1 

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